Current loop as a Magnetic Dipole – Magnetism | Class 12th Physics Notes

Current loop as a Magnetic Dipole

Ampere found that the distribution of magnetic lines of force around a finite current-carrying solenoid is similar to that produced by a bar magnet. This is evident from the fact that a compass needle when moved around these two bodies shows similar deflections. After noting the close resemblance between these two, Ampere demonstrated that a simple current loop behaves like a bar magnet and put forward that all the magnetic phenomena are due to circulating electric current. This is Ampere’s hypothesis.

We consider a circular coil carrying current I. When seen from above current flows in the anti-clockwise direction.

Current loop as a Magnetic Dipole

  1. The magnetic field lines due to each elementary portion of the circular coil are circular near the element and almost straight near the center of the circular coil.

    Current loop as a Magnetic Dipole
  2. The magnetic lines of force seem to enter at the lower face of the coil and leave at the upper face. 
  3. The lower face through which lines of force enter behaves as a south pole and the upper face through which field lines leave behaves as the north pole.
  4. A planar loop of any shape behaves as a magnetic dipole.
  5. The dipole moment of current loop $(\mathrm{M})=$ ampere turns (nI) $\times$ area of coil (A) or $\mathrm{M}=\mathrm{nIA}$.
  6. The unit of dipole moment is ampere meter $^{2}\left( A – m ^{2}\right)$
  7. The magnetic dipole moment is a vector with direction from S pole to N pole or along the direction of normal to the planar area.
 

Atoms as a Magnetic Dipole

In an atom, electrons revolve around the nucleus. These moving electrons behave as small current loops. So atom possesses a magnetic dipole moment and hence behaves as a magnetic dipole.

The angular momentum of electron due to orbital motion $L = m _{ e } vr$

The equivalent current due to orbital motion $I =-\frac{ e }{ T }=-\frac{ ev }{2 \pi r }$



–ve sign shows direction of current is opposite to direction of motion of electron.

Magnetic dipole moment $M=I A=-\frac{e v}{2 \pi r} \cdot \pi r^{2}=-\frac{e v r}{2}$

Using $L=m_{e}$ vr we have $\quad M=-\frac{e}{2 m_{e}} L$

In vector form $\overrightarrow{ M }=-\frac{ e }{2 m _{ e }} \overrightarrow{ L }$

The direction of magnetic dipole moment vector is opposite to angular momentum vector.

According to Bohr’s theory $L=\frac{n h}{2 \pi} \quad n=0, \quad 1, \quad 2 \ldots \ldots$

So $M=\left(\frac{e}{2 m_{e}}\right) \frac{n h}{2 \pi}=n\left(\frac{e h}{4 \pi m_{e}}\right)=n \mu_{B}$

Where $\mu_{\mathrm{B}}=\frac{\mathrm{eh}}{4 \pi \mathrm{m}_{e}}$

$=\frac{\left(1.6 \times 10^{-19} \mathrm{C}\right)\left(6.62 \times 10^{-34} \mathrm{Js}\right)}{4 \times 3.14 \times\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}$

$=9.27 \times 10^{-24} \mathrm{Am}^{2}$

is called Bohr Magneton. This is natural unit of magnetic moment.


  Also Read:

Biot Savart’s Law

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Class 12 Magnetism – Gauss Law Definition || Solved Examples
As you know that the science is filled with fun facts. The deeper one dives into the concepts of science and its related fields, the greater amount of knowledge and information there is to learn in there. One such topic of study is the Gauss Law, which studies electric Charge along with a surface and the topic of Electric Flux. Let us study about the Gauss Law definition , Formula, Solved Examples in this Article,. Gauss’s law states that the net flux of an Electric Field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism. It was initially formulated by Carl Friedrich Gauss in the year 1835 and relates the electric fields at the points on a closed surface and the net charge enclosed by that surface. The surface integral of magnetic field $\overrightarrow{ B }$ over a closed surface S is always zero Mathematically $\oint_{S} \vec{B} \cdot \overrightarrow{d a}=0$
  1. Isolated magnetic poles do not exist is a direct consequence of gauss law in magnetism.
  2. The total magnetic flux linked with a closed surface is always zero.
  3. If a number of magnetic field lines are leaving a closed surface, an equal number of field lines must also be entering the surface.

Ex. A bar magnet of length 0.1 m has a pole strength of 50 Am. Calculate the magnetic field at a distance of 0.2 m from its centre on its equatorial line. Sol. $B _{ equi }=\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}=\frac{10^{-7} \times 50 \times 0.1}{\left(0.2^{2}+0.05^{2}\right)^{\frac{3}{2}}}=\frac{5 \times 10^{-7}}{(0.04+0.0025)^{\frac{3}{2}}}$ or $B _{\text {equi }}=5.7 \times 10^{-5}$ Tesla
Ex. What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5 cm at a distance of 50 cm from its mid-point. The magnetic moment of the bar magnet is 0.40 $Am ^{2}$ Sol. Here r >> $\ell$. So equatorial field $B _{\text {equi }}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}=\frac{10^{-7} \times 0.4}{(0.5)^{3}}=3.2 \times 10^{-7} T$ Axial field $B _{\text {axial }}=\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}=2 \times 3.2 \times 10^{-7}=6.4 \times 10^{-7} T$
Ex. Find the magnetic field due to a dipole of magnetic moment 1.2 $Am ^{2}$ at a point 1 m away from it in a direction making an angle of 60° with the dipole axis? Sol. $B =\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{1+3 \cos ^{2} \theta}=\frac{10^{-7} \times 1.2 \sqrt{1+3 \cos ^{2} 60}}{1}=\frac{10^{-7} \times 1.2 \times \sqrt{7}}{2}=1.59 \times 10^{-7} T$ $\tan \theta^{\prime}=\frac{1}{2} \tan \theta=\frac{1}{2} \tan 60^{\circ}=\frac{\sqrt{3}}{2}=0.866$ So $\theta^{\prime}=\tan ^{-1} 0.866=40.89^{\circ}$
Ex. A circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. Calculate the work required to turn the coil in an external field of 1.5 T through $180^{\circ}$ about an axis perpendicular to the magnetic field. The plane of coil is initially at right angles to magnetic field? Sol. Work done W = MB $\left(\cos \theta_{1}-\cos \theta_{2}\right)=N I A B\left(\cos \theta_{1}-\cos \theta_{2}\right)$ or $W = NI _{ B }^{2} B \left(\cos \theta_{1}-\cos \theta_{2}\right)=100 \times 0.1 \times 3.14 \times(0.05)^{2} \times 1.5\left(\cos 0^{\circ}-\cos \pi\right)=0.2355 J$
Ex. A bar magnet of magnetic moment $1.5 HT ^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required to turn the magnet so as to align its magnetic moment. (i) Normal to the field direction? (ii) Opposite to the field direction? (b) What is the torque on the magnet in case (i) and (ii)? Sol. Here, M = $1.5 JT ^{-1}, B =0.22 T$ (a) P.E. with magnetic moment aligned to field = – MB P.E. with magnetic moment normal to field = 0 P.E. with magnetic moment antiparallel to field = + MB (i) Work done = increase in P.E. = 0 – (–MB) = MB = 1.5 × 0.22 = 0.33 J. (ii) Work done = increase in P.E. = MB – (–MB) = 2MB = 2 × 1.5 × 0.22 = 0.66 J. (b) We have $\tau$ = MB sin $\theta$ (i) $\theta=90^{\circ}, \sin \theta=1, \tau= MB \sin \theta=1.5 \times 0.22 \times 1=0.33 J$ This torque will tend to align M with B. (ii) $\theta=180^{\circ}, \sin \theta=0, \tau= MB \sin \theta=1.5 \times 0.22 \times 0=0$
Ex. A short bar magnet of magnetic moment 0.32 J/T is placed in uniform field of 0.15 T. If the bar is free to rotate in plane of field then which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is potential energy of magnet in each case? Sol. (i) If M is parallel to B then $\theta=0^{\circ}$ So potential energy $U=U_{\min }=-M B$ $U_{\min }=-M B=-0.32 \times 0.15 J=-4.8 \times 10^{-2} J$ This is case of stable equilibrium (ii) If M is antiparallel to B then $\theta=\pi^{\circ}$ so potential energy $U=U_{\max }=+M B=+0.32 \times 0.15=4.8 \times 10^{-2} J$ This is case of unstable equilibrium.
Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Potential Energy of Magnetic Dipole in Magnetic Field || Magnetism Class 12 Physics Notes
Potential Energy of magnetic dipole in a magnetic field is defined as the amount of work done in rotating the dipole from zero potential energy position to any desired position.

A current loop does not experience a net force in a magnetic field. It, however, experiences a torque. This is very similar to the behavior of an electric dipole in an electric field. A current loop, therefore, behaves like a magnetic dipole.

Potential Energy of a Bar Magnet in Uniform Magnetic Field

When a bar magnet of dipole moment M is kept in a uniform magnetic field B it experiences a torque $\tau=M B \sin \theta$ which tries to align it parallel to the direction of the field.

If the magnet is to be rotated against this torque work has to be done.

The work done in rotating dipole by small-angle d$\theta $ is $d W =\tau d \theta$

Total work done in rotating it from angle $\theta_{1}$ to $\theta_{2}$ is

$\mathrm{W}=\int \mathrm{dW}=\int_{\theta_{1}}^{\theta_{2}} \tau \mathrm{d} \theta=\mathrm{MB} \int_{\theta_{1}}^{\theta_{2}} \sin \theta \mathrm{d} \theta$

$=\operatorname{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$

This work done in rotating the magnet is stored inside the magnet as its potential energy.

So U = MB $\left(\cos \theta_{1}-\cos \theta_{2}\right)$

The potential energy of a bar magnet in a magnetic field is defined as work done in rotating it from a direction perpendicular to the field to any given direction.

$U = W _{ \theta }- W _{\frac{\pi}{2}}=- MB \cos \theta=-\overrightarrow{ M } \cdot \overrightarrow{ B }$

Potential Energy of Magnetic Dipole in Magnetic Field

Also Read:

Biot Savart’s Law    
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Magnetic Field due to a Short Bar Magnet | Magnetic Dipole – Class 12 physics Notes
The magnetic field due to a short bar magnet at any point on the axial line is twice the magnetic field at a point on the equatorial line of that magnet at the same distance. S.l. unit of torque acting on the bar magnet is Nm.

Magnetic field due to a short bar magnet (magnetic dipole) :

On Axial Point or End on Position

The magnetic field $\overrightarrow{ B }_{\text {axial }}$ at a point P due to bar magnet will be the resultant of the magnetic fields $\overrightarrow{ B _{1}}$ due to N-pole of magnet and $\overrightarrow{ B _{2}}$ due to S-pole of magnet.

Magnetic field due to a short bar magnet

$\overrightarrow{ B }_{\text {axial }}=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$

$\overrightarrow{ B }_{1}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r -\ell)^{2}}(\hat{ r })$ and $\overrightarrow{ B }_{2}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r +\ell)^{2}}(-\hat{ r })$

$\therefore \quad \vec{B}_{\text {axal }}=\left[\frac{\mu_{0}}{4 \pi} \frac{m}{(r-\ell)^{2}}-\frac{\mu_{0}}{4 \pi} \frac{m}{(r+\ell)^{2}}\right]$

$\hat{\mathrm{r}}=\frac{\mu_{0} \mathrm{~m}}{4 \pi}\left[\frac{4 \mathrm{r} \ell}{(\mathrm{r}-\ell)^{2}(\mathrm{r}+\ell)^{2}}\right] \hat{\mathrm{r}}$

$\overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ Mr }}{\left( r ^{2}-\ell^{2}\right)^{2}}$

Magnetic field due to a bar magnet at an axial point has same direction as that of its magnetic dipole moment vector.

For a bar magnet of very small length $\ell<< r \overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ M }}{r^{3}}$

Browse More Topics Related Magnetism:

On Equatorial Point or Broadside Position 

The magnetic field $\overrightarrow{ B }_{\text {equi }}$ at a point P due to bar magnet will be the resultant of the magnetic fields $\overrightarrow{ B _{1}}$ due to N-pole of magnet and $\overrightarrow{ B _{2}}$ due to S-pole of magnet $\overrightarrow{ B }_{ equi }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$ due to S-pole of magnet $\overrightarrow{ B }_{ equ i }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$

$\left|\overrightarrow{ B }_{1}\right|=\frac{\mu_{0}}{4 \pi} \frac{ m }{ NP ^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ m }{ r ^{2}+\ell^{2}}$ along $NP$

$\left|\overrightarrow{ B }_{2}\right|=\frac{\mu_{0}}{4 \pi} \frac{ m }{ SP ^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)}$ along $PS$



So $\left|\vec{B}_{1}\right|=\left|\vec{B}_{2}\right|$ On resolving $\vec{B}_{1}$ and $\overrightarrow{ B _{2}}$ along PX’ and PY we find $\left|\vec{B}_{1}\right|$ $\sin \theta$ and $|\overrightarrow {{B_2}} |\,\sin \theta $

are equal and opposite so they cancel each other. So resultant field

$\overrightarrow{\mathrm{B}}_{\text {equi }}=\overrightarrow{\mathrm{B}}_{1} \cos \theta(-\hat{\mathrm{r}})+\overrightarrow{\mathrm{B}}_{2} \cos \theta(-\hat{\mathrm{r}})$

$=\left[\frac{\mu_{0}}{4 \pi} \frac{m}{\left(r^{2}+\ell^{2}\right)} \cos \theta+\frac{\mu_{0}}{4 \pi} \frac{m}{\left(r^{2}+\ell^{2}\right)} \cos \theta\right](-\hat{r})$

$=2 \cdot \frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cdot \frac{\ell}{\sqrt{ r ^{2}+\ell^{2}}}(-\hat{ r })$

$\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$

$\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$

The direction of magnetic field at a point on equitorial line is opposite to magnetic dipole moment vector.

For a bar magnet of very small length $\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{r^{3}}$

At An Arbitrary Point

The point P is on axial line of magnet S’N’ with magnetic moment Mcos$\theta $ Magnetic flux density $B _{1}=\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{ r ^{3}}$

The point P is simultaneously on the equatorial line of other magnet N”S” with magnetic moment Msin $\theta $ Magnetic flux density $B _{2}=\frac{\mu_{0}}{4 \pi} \frac{ M \sin \theta}{ r ^{3}}$

Total magnetic flux density at P.

$B =\sqrt{ B _{1}^{2}+ B _{2}^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{4 \cos ^{2} \theta+\sin ^{2} \theta}$ or

$\tan \theta^{\prime}=\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{r^{3}}}{\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}}}=\frac{1}{2} \tan \theta$ or $\theta^{\prime}=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)$



 

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Magnetic Field | Properties of Magnetic Lines of Force Class 12, JEE

Magnetic Field

The space around a magnet (or a current-carrying conductor) in which its magnetic effect can be experienced. Magnetic Lines of Force can be defined as curved lines used to represent a magnetic field, drawn such that the number of lines relates to the magnetic field’s strength at a given point and the tangent of any curve at a particular point is along the direction of the magnetic force at that point. The Properties of Magnetic Lines of Force are also discussed.

Magnetic Field:

Magnetic Field is defined as the space around a magnet (or a current-carrying conductor) in which its magnetic effect can be experienced.
  1. The magnetic field in a region is said to be uniform if the magnitude of its strength and direction is the same at all points in that region.
  2. A magnetic field in a region is said to be uniform if the magnitude of its strength and direction is the same at all the points in that region.
  3. The strength of the magnetic field is also known as magnetic induction or magnetic flux density.
  4. The $\mathrm{SI}$ unit of the strength of the magnetic field is Tesla (T)1 Tesla = 1 newton ampere $^{-1}$ metre $^{-1}\left( NA ^{-1} m ^{-1}\right)=1$ Weber metre $^{-2}\left( Wb m ^{-2}\right)$
  5. The CGS unit is Gauss (G)1 Gauss (G) $=10^{-4}$ Tesla (T)

Properties of Magnetic Lines of Force:

The magnetic field lines are the graphical method of representation of the magnetic field. This was introduced by Michael Faraday.
  1.  A line of force is an imaginary curve the tangent to which at a point gives the direction of the magnetic field at that point
  2. The magnetic field line is the imaginary path along which an isolated north pole will tend to move if it is free to do so.
  3. The magnetic lines of force are closed curves. They appear to converge or diverge at poles. outside the magnet, they run from north to south pole and inside from south to north.
  4. The number of lines originating or terminating on a pole is proportional to its pole strength.Magnetic flux = number of magnetic lines of force = $\mu_{0} \times m$ Where $${\mu _0}$$ is a number of lines associated with a unit pole.
  5. Magnetic lines of force do not intersect each other because if they do there will be two directions of the magnetic field which is not possible.
  6. The magnetic lines of force may enter or come out of the surface at any angle.
  7. The number of lines of force per unit area at a point gives the magnitude of the field at that point. The crowded lines show a strong field while distant lines represent a weak field.
  8. The magnetic lines of force have a tendency to contract longitudinally like a stretched elastic string producing attraction between opposite poles.
  9. The magnetic lines of force have a tendency to repel each other laterally resulting in repulsion between similar poles.
  10. The region of space with no magnetic field has no lines of force. At the neutral point where the resultant field is zero, there cannot be any line of force.
  11. Magnetic lines of force exist inside every magnetized material.                                           
Important points :
  1. Magnetic lines of force always form closed and continuous curves whereas the electric lines of force are discontinuous.
  2. Each electric line of force starts from a positive charge and ends at a negative charge. Electric lines of force are discontinuous because no such lines exist inside a charged body.
  3. In magnetism, as there are no monopolies, therefore, the magnetic field lines will be along with closed loops with no starting or ending. The magnetic lines of force would pass through the body of the magnet.
  4. At very far-off points, the lines due to an electric dipole and a magnetic dipole appear identical.
 
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Magnetic Dipole Moment Definition, Formulas & Solved Examples | Class 12, JEE & NEET
A magnetic moment is a quantity that represents the magnetic strength and orientation of a magnet or any other object that produces a magnetic field. More precisely, a magnetic moment refers to a magnetic dipole moment, the component of the magnetic moment that can be represented by a magnetic dipole. Know the Magnetic Dipole Moment Definition, Formulae, and solved examples here.

Magnetic Dipole:

An arrangement of two magnetic poles of equal and opposite strengths separated by a finite distance is called a magnetic dipole.

 
  1. Two poles of a magnetic dipole or a magnet are of equal strength and opposite nature.

  2. The line joining the poles of the magnet is called the magnetic axis.

  3. The distance between the two poles of a bar magnet is called the magnetic length of the magnet. It is denoted by 2$\ell$

  4. The distance between the ends of the magnet is called the geometrical length of the magnet.

  5. The ratio of magnetic length and geometrical length is $\frac{5}{6}$ or 0.83

  6. A small bar magnet is treated like a magnetic dipole.

Magnetic Dipole Moment Definition:

The product of strength of either pole and the magnetic length of the magnet is called a magnetic dipole moment. $\overrightarrow{ M }= m (\overrightarrow{2 \ell})$  

Important Points to Remember


    1. It is a vector quantity whose direction is from the south pole to the north pole of a magnet.

    2. The unit of magnetic dipole moment is ampere metre $^{2}\left( Am ^{2}\right)$ and Joule/Tesla (J/T). The dimensions are $M^{0} L^{2} T^{0} A^{1}$ 

    3. If a magnet is cut into two equal parts along the length then pole strength is reduced to half and length remains unchanged.New magnetic dipole moment M’ = m’ $(2 \ell)=\frac{ m }{2} \times 2 \ell=\frac{ M }{2}$The new magnetic dipole moment of each part becomes half of the original value.
 
  1. If a magnet is cut into two equal parts transverse to the length then pole strength remains unchanged and length is reduced tohalf. New magnetic dipole moment $M^{\prime}=m\left(\frac{2 \ell}{2}\right)=\frac{M}{2}$
    The new magnetic dipole moment of each part becomes half of the original value.

  2. In magnetism existence of magnetic monopole is not possible.

  3. The magnetic dipole moment of a magnet is equal to the product of pole strength and distance between poles. M = m d

  4. As the magnetic moment is a vector, in the case of two magnets having magnetic moments $M_{1}$ and

    $M_{2}$
    with angle

    $\theta$ between them, the resulting magnetic moment.
    $M =\left[ M _{1}^{2}+ M _{2}^{2}+2 M _{1} M _{2} \cos \theta\right]^{1 / 2}$ with $\tan \phi=\left[\frac{M_{2} \sin \theta}{M_{1}+M_{2} \cos \theta}\right]$

Ex. The force between two magnetic poles in air is 9.604 mN. If one pole is 10 times stronger than the other, calculate the pole strength of each if distance between two poles is 0.1 m? Sol. Force between poles $F =\frac{\mu_{0}}{4 \pi} \frac{ m _{1} m _{2}}{ r ^{2}}$ or $9.604 \times 10^{-3}=\frac{10^{-7} \times m \times 10 m}{0.1 \times 0.1}$ or $m ^{2}=96.04 N ^{2} T ^{-2}$ or m = 9.8 N/T So strength of other pole is 9.8 × 10 = 98 N/T
Ex. A steel wire of length L has a magnetic moment M. It is then bent into a semicircular arc. What is the new magnetic moment? Sol. If m is the pole strength then M = m . L or $m =\frac{ M }{ L }$ If it is bent into a semicircular arc then L = $\pi$ or $r=\frac{L}{\pi}$ So new magnetic moment $M^{\prime}=m \times 2 r=\frac{M}{L} \times 2 \times \frac{L}{\pi}=\frac{2 M}{\pi}$
Ex. Two identical bar magnets each of length L and pole strength m are placed at right angles to each other with the north pole of one touching the south pole of other. Evaluate the magnetic moment of the system. Sol. $M _{1}= M _{2}= mL$
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Coulomb’s law of Magnetism || Magnetism and Matter Class 12, JEE & NEET
Magnetism and Matter Class 12 Notes will help you in your Exam Preparation and will also help in scoring good! The property of any object by virtue of which it can attract a piece of iron or steel is called Magnetism. Here we will study about the Coulomb’s Law in Magnetism, Magnetic Flux Density, and Pole Strength.
  1. Coulomb’s Law in Magnetism
  2. Magnetic Flux Density
  3. Pole Strength

Coulomb’s Law in Magnetism

Coulomb's law of Magnetism

If two magnetic poles of strengths ${{m_1}}$ and ${{m_2}}$ are kept at a distance r apart then the force of attraction or repulsion between the two poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them

$$F \propto {{{m_1}{m_2}} \over {{r^2}}}\quad or\,\,F = {{{\mu _0}} \over {4\pi }}{{{m_1}{m_2}} \over {{r^2}}}$$

Where $\frac{\mu_{0}}{4 \pi}=10^{-7} Wb A ^{-1} m ^{-1}=10^{-7}$ henry/m where $\mu_{0}$ is permeability of free space.

Magnetic Flux Density

Coulomb's law of Magnetism
The force experienced by a unit north pole when placed in a magnetic field is called magnetic flux density or field intensity at that point

$\overrightarrow{ B }=\frac{\overrightarrow{ F }}{ m }=\frac{\mu_{0}}{4 \pi} \frac{ m }{ r ^{2}} \hat{ r }$

This is the magnetic field produced by a pole of strength m at distance r.

Pole Strength

In relation $F=\frac{\mu_{0}}{4 \pi} \frac{m_{1} m_{2}}{r^{2}}$

If $m _{1}= m _{2}= m , r =1 m$ and $F =10^{-7} N$

Then $10^{-7}=10^{-7} \times \frac{ m \times m }{1^{2}}$ or $m^{2}=1$ or $m=\pm 1$ ampere metre (A-m)

The strength of a magnetic pole is said to be one ampere meter if it repels an equal and similar pole with a force of $10^{-7}$ N when placed in vacuum (or air) at a distance of one meter from it.

The pole strength of the north pole is defined as the force experienced by the pole when kept in a unit magnetic field.

$m =\frac{\overrightarrow{ F }}{\overrightarrow{ B }}$
  1. Pole strength is a scalar quantity with dimension $M^{0} L^{1} T^{0} A^{1}$ 
  2. The unit is newton/Tesla or ampere meter.
  3. The pole strength depends on the nature of the material of the magnet, the state of magnetization (with an upper limit called saturation), and the area of cross-section.
  4. The north pole experiences a force in the direction of the magnetic field while the south pole experiences force opposite to the field.
     
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ATOMIC THEORY OF MAGNETISM || Magnetism and Matter Class 12, JEE & NEET
Magnetism and Matter Class 12 Notes will help you in your Exam Preparation and will also help in scoring good! The property of any object by virtue of which it can attract a piece of iron or steel is called magnetism. Here will study about the Atomic Theory of Magnetism. Atomic Theory of Magnetism :
  1. Each atom behaves like a complete magnet having a north and south pole of equal strength. The electrons revolving around the nucleus in an atom are equivalent to small current loops which behave as a magnetic dipole.
  2. In unmagnetized magnetic substance these atomic magnets (represented by arrows) are randomly oriented and form closed chains. The atomic magnets cancel the effect of each other and thus resultant magnetism is zero.
  3. In a magnetized substance, all the atomic magnets are aligned in the same direction and thus resultant magnetism is non-zero.

The atomic theory explains the following facts in magnetism.

  1. Non existence of monopoles. The magnetic poles always exist in pairs and are of equal strength.
  2. When a magnet breaks, then each part behaves like a complete magnet.
  3. The magnetization of an electromagnet can be explained as the alignment of atomic magnets in direction of the magnetic field.
  4. This explains the phenomenon of saturation magnetization i.e. acquired magnetism remains constant even on increasing the external magnetizing field.
   
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Properties of Bar Magnet || Magnetism and Matter Class 12, JEE & NEET
Playing with magnets is one of the first moments of science most children discover. That’s because magnets are easy to use, safe, and fun. They’re also quite surprising. Remember when you first discovered that two magnets could snap together and stick like glue? Remember the force when you held two magnets close and felt them either attract (pull toward one another) or repel (push away)? Here we will study about the Bar Magnet and Properties of Bar Magnet! A bar magnet is a rectangular piece of the object. It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent Magnetic Properties. The magnet has two poles: a north and a south pole. When you suspend it freely, the magnet aligns itself so that the north pole points towards the magnetic north pole of the earth.

Properties of Bar Magnet :

  1. Attractive Property and Poles : When a magnet is dipped into iron fillings it is found that the concentration of iron filings, i.e., attracting power of the magnet is maximum at two points near the ends and minimum at the center. The places in a magnet where its attracting power is maximum are called poles while the place of minimum attracting power is called the neutral region.
  2. Directive Property and N-S Poles: When the magnet is suspended its length becomes parallel to the N-S direction. The pole pointing north is called the north pole while the other pointing south is called the south pole.
  3. Magnetic Axis and Magnetic Meridian: The line joining the two poles of a magnet is called the magnetic axis and the vertical plane passing through the axis of a freely suspended or pivoted magnet is called a magnetic meridian.
  4. Magnetic Length: The distance between two poles along the axis of a magnet is called its effective or magnetic length. As poles are not exactly at the ends, the effective length is lesser than the actual length of the magnet.
  5. Poles Exist in Paris: In a magnet, the two poles are found to be equal in strength and opposite in nature. If a magnet is broken into a number of pieces, each piece becomes a magnet with two equal and opposite poles. This shows that monopoles do not exist.
  6. Consequent-poles and No-pole: Monopoles do not exist in a magnet but there are two poles of equal strength and opposite nature : (a) There can be magnets with no poles, e.g., a magnetized ring called toroid or solenoid of infinite length has properties of a magnet but no poles.

    (b) There can be magnets with two similar poles (or with three poles), e.g., due to faulty magnetization of a bar, temporarily identical poles at the two ends with an opposite pole of double strength at the center of the bar (called consequent pole) are developed.
  7. Repulsion is a Sure Test of Polarity: A pole of a magnet attracts the opposite pole while repels a similar pole. A sure test of polarity is repulsion and not attraction, as attraction can take place between opposite poles or a pole and a piece of an unmagnetized magnetic material due to the ‘induction effect’.
  8. Magnetic Induction: A magnet attracts certain other substances through the phenomenon of magnetic induction i.e., by inducing the opposite pole in magnetic material on the side facing it as shown in fig.
  9. Magnetic and Non-magnetic Materials: The substances such as steel, iron, cobalt, and nickel, etc., which are attracted by a magnet are called magnetic while substances such as copper, aluminum stainless steel, wood, glass, and plastic, etc. which are not attracted by the magnet are usually called non-magnetic.
  10. Permanent and Temporary Magnets: If a magnet retains its attracting power for a long time it is said to be permanent, otherwise temporary. Permanent magnets are made of steel, Alnico, Alcomax, or Ticonal while temporary of soft iron, mumetal, or stalloy.
  11. Demagnetization: A magnet gets demagnetized, i.e., loses its power of attraction if it is heated, hammered or ac is passed through a wire wound over it.
  12. Magnetic Keepers: A magnet tends to become weaker with age owing to self-demagnetization due to poles at the ends which tend to neutralize each other. However, by using pieces of soft iron called keepers, the poles at the ends are neutralized and consequently, the demagnetizing effect disappears and the magnet can retain its magnetism for a longer period.
                                                       

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Magnetism and Matter Class 12 Notes – Introduction
Magnetism and Matter Class 12 Notes will help you in your Exam Preparation and will also help in scoring good! The property of any object by virtue of which it can attract a piece of iron or steel is called magnetism. Natural Magnet A natural magnet is an ore of iron (Fe3O4), which attracts small pieces of iron, cobalt and nickel towards it. Magnetite or lode stone is a natural magnet. Artificial Magnet A magnet which is prepared artificially is called an artificial magnet, e.g., a bar magnet, an electromagnet, a magnetic needle, a horse-shoe magnet etc. According to molecular theory, every molecular of magnetic substance (whether magnetised or not) is a complete magnet itself. The phenomenon of attracting magnetic substances like iron, cobalt nickel etc is called magnetism. A body possessing the property of magnetism is called magnet. Historical facts : (1) The word magnet is derived from the name of an island in Greece called magnesia where magnetic ore deposits were found. (2) Thales of Miletus knew that pieces of lodestone or magnetite (black iron oxide $Fe _{2} O _{3}$) could attract small pieces of iron. (3) The Chinese discovered that a linear piece of lodestone when suspended freely pointed in north and south direction. That is why name lodestone which is given to magnetite means leading stone. (4) The Chinese are credited with making technological use of this directional property for navigation of ships. (5) In 1600 BC William Gilbert published a book De Magnete which gave an account of then known facts of magnetism. (6) Due to their irregular shapes and weak attracting power natural magnets are rarely used. (7) Lodestone or magnetite is naaaatural magnet. Earth is also a natural magnet. Artificial magnets :
  1. The permanent artificial magnets are made of some metals and alloys like carbon-steel, Alnico, Platinum-cobalt, Alcomax, Ticonal. The permanent magnets are made of ferromagnetic substances with large coercivity and retentivity and can have desired shape like bar-magnet, U shaped magnet or magnetic needle etc. These magnet retain its attracting power for a long time.
2. The temporary artificial magnet like electromagnets are prepared by passing current through coil wound on soft iron core. These cannot retain its attracting power for a long time.  
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Construction and Working Principle of Cyclotron Class 12 Physics
We will study here about the Construction and Working Principle of Cyclotron Class 12
  1. Working Principle of Cyclotron
  2. Construction of Cyclotron

Cyclotron Introduction

A cyclotron is used for accelerating positive ions so that they acquire energy large enough to carry out nuclear reactions.
Principle of Cyclotron Class 12 Cyclotron was designed by Lawrence and Livingstone in 1931.

In a cyclotron, the positive ions cross again and again the same alternating (radio frequency) electric field.

And gain the energy each time = q V.

q = charge and $V=p o t^{n}$ . difference in betn dees.

It is achieved by making them to move along the spiral paths under the action of a strong magnetic field.

Working Principle of Cyclotron

A positive ion can acquire sufficiently large energy with a comparatively smaller alternating potential difference by making them to cross the same electric field again and again by making use of a strong magnetic field.

Construction of Cyclotron

It consists of two D-shaped hollow semicircular metal chambers $\mathrm{D}_{1}$ and $\mathrm{D}_{2}$, called dees.

The two dees are placed horizontally with a small gap separating them.

The dees are connected to the source of the high-frequency electric field.

The dees are enclosed in a metal box containing a gas at a low pressure of the order of

10–3 mm mercury.

The whole apparatus is placed between the two poles of a strong electromagnet NS as shown in fig. The magnetic field acts perpendicular to the plane of the dees.

The positive ions are produced in the gap between the two dees by the ionization of the gas.

To produce a proton, hydrogen gas is used; while for producing $\alpha$ -particles, helium gas is used. Theory: Consider that a positive ion is produced at the center of the gap at the time, when the dee $D_{1}$ is at a positive potential and the dee $D_{2}$, is at a negative potential. The positive ion will move from dee $D_{1}$ to dee $D_{2}$ The force on the positive ion due to the magnetic field provides the centripetal force to the positive ion and it is deflected along a circular path because the magnetic field is normal to the motion. Let strength of the magnetic field $=\mathrm{B}$ mass of ion $=\mathrm{m}, \quad$ velocity of ion $=\mathrm{v} \quad$ and $\quad$ charge of the positive ion $=\mathrm{q}$ and the radius of the semi-circular path $=\mathrm{r}$

then $\left.\quad \mathrm{Bqv}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \quad \text { [inside the dee } \mathrm{D}_{2}\right]$

Thus, $r=\frac{m v}{B q}$

After moving along the semi-circular path inside the dee $D_{2},$ the positive ion reaches the gap between the dees.

At this stage, the polarity of the dees just reverses due to alternating “electric field” i.e. dee $D_{1}$, becomes negative and dee $\mathrm{D}_{2}$ becomes positive. The positive ion again gains the energy, as it is attracted by the dee $D_{1}$, After moving along the semi-circular path inside the dee $D_{1}$, the positive ion again reaches the gap and it gains the energy. ( $=\mathrm{q} \mathrm{V}$ ) This process repeats itself because the positive ion spends the same time inside a dee irrespective of its velocity or the radius of the circular path.

The time spent inside a dee to cover semi-circular path,

is $\quad \mathrm{t}=\frac{\text { length of the semi circular path }}{\text { velocity }}=\frac{\pi \mathrm{r}}{\mathrm{v}}$

Or $\mathrm{t}=\frac{\pi \mathrm{m}}{\mathrm{Bq}} \quad\left[\frac{\mathrm{r}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{Bq}}\right]$

As positive ion gains kinetic energy its velocity increases, due to increasing velocity, decrease in time spent inside a dee of positive ions is exactly compensated by the increase in length of the semi circular path (r $\propto$ v).

Due to this condition, the positive ion always crosses the alternating electric field across the gap incorrect phase.

 

    Also Read:

 

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Force Between Two Parallel Current Carrying Conductors || Class 12 Physics Notes
It is an experimentally established fact that two current-carrying conductors attract each other when the current is in the same direction and repel each other when the currents are in opposite direction. Here we will study about Force Between Two Parallel Current Carrying Conductors as wire:

Force Between Parallel Current-Carrying Wires

Consider two long wires $W_{1}$ and $W_{2}$ kept parallel to each other and carrying currents $I_{1}$ and $\mathrm{I}_{2}$ respectively in the same direction. The separation between the wires is d. Consider a small element d\ell of the wire $WA_{2}$ The magnetic field at d\ell due to the wire $W_{1}$ is

Force Between Two Parallel Current Carrying Conductors$B_{1}=\frac{\mu_{0} I_{1}}{2 \pi d}$ …….(i)

The field due to the portions of the wire $W_{2},$ above and below $d \ell,$ is zero. Thus, eq” (i) gives

the net field at $\mathrm{d} \ell$ . The direction of this field is perpendicular to the plane of the diagram and going into it. The magnetic force at the element $d \ell$ due to wire $w_{1}$ is.

The vector product $\overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{B}}$ has a direction towards the wire $\mathrm{W}_{1} .$ Thus, the length $\mathrm{d} \ell$ of wire $\mathrm{W}_{2}$ is attracted towards the wire $\mathrm{W}_{1}$. The force per unit length of the wire $\mathrm{W}_{2}$ due to the wire $W_{1}$ is

  If we take an element $\mathrm{d} \ell$ in the wire $\mathrm{W}_{1}$ and calculate the magnetic force per unit length on wire $W_{1}$ due to $\mathrm{W}_{2},$ it is again given by $\operatorname{eq}^{n}(i i)$

Force Between Two Parallel Current Carrying Conductors If the parallel wires currents in opposite directions, the wires repel each other.

The wires attract each other if the current in the wires is flowing in the same direction.

And they repel each other if the currents are in opposite directions.

 

Experimental Demonstration

Force Between Two Parallel Current Carrying Conductors Definition of Ampere

$\because \quad \mathrm{F}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}} \mathrm{N} / \mathrm{m}$

When $\mathrm{I}_{1}=\mathrm{I}_{2}=1$ ampere and $\mathrm{r}=1 \mathrm{m},$ then $\quad \mathrm{F}=\frac{\mu_{0}}{2 \pi}=\frac{4 \pi \times 10^{-7}}{2 \pi} \mathrm{N} / \mathrm{m}=2 \times 10^{-7} \mathrm{N} / \mathrm{m}$ This leads to the following definition of ampere.

One ampere is that current which, if passed in each of two parallel conductors of infinite length and one metre apart in vacuum causes each conductor to experience a force of $2 \times 10^{-7}$ newton per metre of length of conductor.

Dimensional of formula of $\mu_{0}$

$\because \quad \mathrm{F}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}} \quad$ so

$\left[\mu_{0}\right]=\frac{[\mathrm{F}][\mathrm{r}]}{\left[\mathrm{I}_{1} \mathrm{I}_{2}\right]}=\frac{\left[\mathrm{ML}^{0} \mathrm{T}^{-2}\right][\mathrm{L}]}{\left[\mathrm{I}^{2}\right]}=\left[\mathrm{MLT}^{-2} \mathrm{I}^{-2}\right]$

  Also Read: Biot Savart’s Law  

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Force on a Moving Charge in a Magnetic Field | Class 12 Physics Notes
  1. Force on a Charged Particle in a Magnetic Field
  2. Difference in Force on a Charged Particle by Magnetic Field and Electric Field
When a charged particle travels through a magnetic field, it experiences a force unlike any other that we’re familiar with in everyday life. To illustrate the point, envision yourself walking down the sidewalk, when all of a sudden, a strong gust of wind hits you from the side. Now imagine that instead of moving sideways, you shoot straight up to the sky. Here we will study about the Force on a Moving Charge in a Magnetic Field.

Force on a Charged Particle in a Magnetic Field

Force experienced by a current element Id $\vec{\ell}$ in magnetic field $\overrightarrow{\mathrm{B}}$ is given by

$\mathrm{dF}=1 \mathrm{~d} \vec{\ell} \times \overrightarrow{\mathrm{B}}$ ………(1)

Now if the current element $\mathrm{Id} \vec{\ell}$ is due to the motion of charge particles, each particle having a charge q moving with velocity $\overrightarrow{\mathrm{v}}$ through a cross-section S,

$\mathrm{Id} \vec{\ell}=\mathrm{n} \mathrm{S} \mathrm{q} \quad \overrightarrow{\mathrm{v}} \cdot \mathrm{d} \ell=\mathrm{n} \mathrm{d} \tau \mathrm{q} \overrightarrow{\mathrm{v}}$ [with volume $\mathrm{d} \tau=\mathrm{S} \mathrm{d} \ell]$

From eq $^{n}$ (i) we can write $\mathrm{dF}=\mathrm{n} \mathrm{d} \tau \mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$

$n \mathrm{d} \tau=$ the total number of charged particles in volume d\tau $(n=$ number of charged particles per unit volume),

force on a charged particle From this it is clear that : $\left.\vec{F}=\frac{1}{n} \frac{d \vec{F}}{d \tau}=q \quad \vec{v} \times \vec{B}\right)$

(a) The force $\overrightarrow{\mathrm{F}}$ is always perpendicular to both the velocity $\overrightarrow{\mathrm{v}}$ and the field $\overrightarrow{\mathrm{B}}$.

(b) A charged particle at rest in a steady magnetic field does not experience any force.

If the charged particle is at rest then $\overrightarrow{\mathrm{v}}=0,$ so $\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}=0$

(c) A moving charged particle does not experience any force in a magnetic field if its motion is parallel or antiparallel to the field.

i.e., if $\quad \theta=0^{\circ}$ or $180^{\circ}$

Force on a Moving Charge in a Magnetic Field

(d) If the particle is moving perpendicular to the field.

In this situation all the three vectors $\overrightarrow{\mathrm{F}}, \overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{B}}$ are mutually perpendicular to each other. Then $\sin \theta=\max =1,$ i.e., $\theta=90^{\circ}$

The force will be maximum $F_{\max }=q \vee B$

(e) Work done by force due to magnetic field in motion of a charged particle is always zero.

When a charged particle move in a magnetic field, then force acts on it is always perpendicular to displacement,

so the work done, $\left.\quad \mathrm{W}=\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{ds}}=\int \mathrm{F} d \mathrm{s} \cos 90^{\circ}=0 \quad \text { (as } \theta=90^{\circ}\right)$

And as by work-energy theorem $\mathrm{W}=\Delta \mathrm{KE},$ the kinetic energy $\left(=\frac{1}{2} \mathrm{mv}^{2}\right)$, remains unchanged and hence speed of charged particle v remains constant.

However, in this situation the force changes the direction of motion, so the direction of velocity of $\vec{v}$ the charged particle changes continuously.

(f) For motion of charged particle in a magnetic field $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$

So magnetic induction $\overrightarrow{\mathrm{B}}$ can be defined as a vector having the direction in which a moving charged particle does not experience any force in the field and magnitude equal to the ratio of the magnitude of maximum force to the product of the magnitude of charge with velocity

Difference in Force on a Charged Particle by Magnetic Field and Electric Field


Difference in Force on a Charged Particle by Magnetic Field and Electric Field

Also Read: Biot Savart’s Law

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Motion of Charged Particle in a Magnetic Field | Moving Charges and Magnetism Class 12, JEE & NEET
When a charged particle moves along a magnetic field line into a region where the field becomes stronger, the particle experiences a force that reduces the component of velocity parallel to the field. This force slows the motion along the field line and here reverses it, forming a Magnetic Mirror. The motion of a charged particle in a magnetic field is characterized by the change in the direction of motion. It is expected also as the magnetic field is capable of only changing the direction of motion. In order to keep the context of the study simplified, we assume the magnetic field to be uniform. This assumption greatly simplifies the description and lets us easily visualize the motion of a charged particle in a magnetic field.

Motion of a Charged Particle in a Magnetic Field

The motion of a charged particle when it is moving collinear with the field, the magnetic field is not affected by the field (i.e. if motion is just along or opposite to a magnetic field) ( $ \quad F=0$ ) Only the following two cases are possible:

The case I: When the charged particle is moving perpendicular to the field The angle between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{v}}$ is $\theta=90^{\circ}$

So the force will be maximum ( $=$ qvB ) and always perpendicular to motion (and also field);

Hence the charged particle will move along a circular path (with its plane perpendicular to the field).

Centripetal force is provided by the force qvB,

Motion of a charged particle in magnetic field

In case of the circular motion of a charged particle in a steady magnetic field :



i.e., with the increase in speed or kinetic energy, the radius of the orbit increases. For uniform circular motion $v=\omega r$

Angular frequency of circular motion called cyclotron or gyro-frequency. $\omega=\frac{\mathrm{v}}{\mathrm{r}}=\frac{\mathrm{qB}}{\mathrm{m}}$

and the time period, $\quad \mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \frac{\mathrm{m}}{\mathrm{qB}}$

i.e., time period (or frequency) is independent of speed of particle and radius of the orbit.

Time period depends only on the field B and the nature of the particle,

i.e., specific charge (q/m) of the particle.

This principle has been used in a large number of devices such as cyclotron (a particle accelerator), bubble-chamber (a particle detector) or mass-spectrometer etc.

the motion of a charged particle in an electric and magnetic field

 

case II : The charged particle is moving at an angle $\theta$ to the field :

$\left(\theta \neq 0^{\circ}, 90^{\circ} \text { or } 180^{\circ}\right)$

Resolving the velocity of the particle along and perpendicular to the field.

The particle moves with constant velocity v cos $\theta$ along the field ($\because$ no force acts on a charged particle when it moves parallel to the field).

And at the same time, it is also moving with velocity $v$ sin $\theta$ perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field)





Motion of a charged particle in magnetic field

So the resultant path will be a helix with its axis parallel to the field $\overrightarrow{\mathrm{B}}$ as shown in fig. The pitch p of the helix $=$ linear distance travelled in one rotation

$p=T(v \cos \theta)=\frac{2 \pi m}{q B}(v \cos \theta)$

  Also Read:

Biot Savart’s Law

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Torque on a Current loop in a Magnetic Field || Moving Charges and Magnetism – Class 12, JEE & NEET
Without torque, there would be no twists and turns, no spins! Wouldn’t life be boring that way? Torque gives a rotational motion to an object that would otherwise not be possible. So here we will study about Torque and Torque on a Current loop in a Magnetic Field in detailed derivation.

Torque Experienced by a Current Loop in a Uniform Magnetic Field

Let us consider a rectangular current loop PQRS of sides a and b suspended vertically in a uniform magnetic field. $\overrightarrow{\mathrm{B}}$. Let $\theta$ be the angle between the direction of $\overrightarrow{\mathrm{B}}$ and the vector perpendicular to the plane of the loop.

Let us first consider the two straight parts PQ and RS.

The forces acting on these parts are clearly equal in magnitude and opposite in direction. These forces are collinear, so, the net force or net torque due to this pair of forces is zero.

The forces on the straight current segments SP and QR are

(i) equal in magnitude (ii) opposite in direction (iii) not collinear.

These forces are shown in figure (b).

This pair of forces produce a torque whose lever arm is b sin\theta. Magnitude of each force $=\mathrm{B} \mathrm{I}$ a

Torque on a Current loop in a Magnetic Field

The magnitude of the torque $\vec{\tau}$ is given by



where $A$ is the area of the coil.

But $\mathrm{IA}=\mathrm{M}$ (magnitude of magnetic dipole moment)



The direction of the torque is vertically downwards along the axis of suspension [dotted line in fig (b)]

It will rotate the loop clockwise about its axis.

Note 1. If the rectangular loop has N turns, then the torque increases N times and becomes $\mathrm{N} \mathrm{B}$ I $\mathrm{A} \sin \theta$

 

Note 2

$\mathrm{eq}^{\mathrm{n}}$ (i) could also be written as $\vec{\tau}=\mathrm{I}(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})$ OR $\vec{\tau}=\mathrm{I} \mathrm{A} \hat{\mathrm{n}} \times \overrightarrow{\mathrm{B}}$

where $\hat{\mathrm{n}}$ is a unit vector normal to the plane of the loop.

Also Read:

Biot Savart’s Law

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Flemings Left Hand Rule Class 12 Physics – eSaral
Flemings Left Hand Rule Class 12 states that if we stretch the thumb, the forefinger, and the middle finger of our left hand such that they are mutually perpendicular to each other. If the forefinger gives the direction of current and the middle finger points in the direction of the magnetic field then the thumb points towards the direction of the force or motion of the conductor.

Flemings Left Hand Rule Class

Then if the forefinger points in the direction of field $(\overrightarrow{\mathrm{B}})$, the central finger in the direction of current, the thumb will point in the direction of the force.

(ii) Right-hand Palm Rule: Stretch the fingers and thumb of the right hand at right angles to each other. Then if the fingers point in the direction of field $\overrightarrow{\mathrm{B}}$ and thumb in the direction of current $\mathbf{I}$, the normal to palm will point in the direction of force.

Right hand Palm rule Regarding the force on a current-carrying conductor in a magnetic field it is worth mentioning that :

(a) As the force BI d\ell sin \theta is not a function of position r, the magnetic force on a current element is non-central [a central force is of the form $\mathrm{F}=\mathrm{Kf}(\mathrm{r}) \overrightarrow{\mathrm{n}_{\mathrm{r}}}]$

(b) The force d\vec $\overrightarrow{\mathrm{F}}$ is always perpendicular to both $\overrightarrow{\mathrm{B}}$ and $\mathrm{Id} \vec{\ell}$ though $\overrightarrow{\mathrm{B}}$ and $\mathrm{Id} \vec{\ell}$ may or may not be perpendicular to each other.

(c) In case of current-carrying conductor in a magnetic field if the field is uniform i.e.,

$\overrightarrow{\mathrm{B}}=$ constt.,

$\overrightarrow{\mathrm{F}}=\int \mathrm{I} \mathrm{d} \vec{\ell} \times \overrightarrow{\mathrm{B}}=\mathrm{I}\left[\int \mathrm{d} \vec{\ell}\right] \times \overrightarrow{\mathrm{B}}$

and as for a conductor $\int_{\mathrm{d}} \vec{\ell}$ represents the vector sum of all the length elements from initial to final point, which in accordance with the law of vector addition is equal to the length vector $\overrightarrow{\ell^{\prime}}$ joining initial to final point, so a current-carrying conductor of any arbitrary shape in a uniform field experience a force

$\vec{F}=I\left[\int d \vec{\ell}\right] \times \vec{B}=I \ell^{\prime} \times \vec{B}$

Flemings Left Hand Rule Class where $\vec{\ell}$ is the length vector joining initial and final points of the conductor as shown in fig.

(d) If the current-carrying conductor in the form of a loop of any arbitrary shape is placed in a uniform field,

$\overrightarrow{\mathrm{F}}=\oint \mathrm{Id\vec{\ell }} \times \overrightarrow{\mathrm{B}}=\mathrm{I}[\oint \overrightarrow{\mathrm{d} \vec{\ell}}] \times \overrightarrow{\mathrm{B}}$

and as for a closed loop, $\oint \mathrm{d} \vec{\ell} \text { is always zero. [vector sum of all } \mathrm{d} \vec{\ell}]$



i.e., the net magnetic force on a current loop in a uniform magnetic field is always zero as shown in fig.

Here it must be kept in mind that in this situation different parts of the loop may experience an elemental force due to which the loop may be under tension or may experience a torque as shown in fig.

Current Loop in a Uniform Field

(e) if a current-carrying conductor is situated in a non-uniform field, its different elements will experience different forces; so in this situation,

$\overrightarrow{\mathrm{F}_{\mathrm{R}}} \neq 0$ but $\overrightarrow{\mathrm{\tau}}$ may or may not be zero




Click here for the Video tutorials of Magnetic Effect of Current Class 12

So, that’s all from this blog. I hope you get the idea about the Flemings Left Hand Rule of Class 12. If you enjoyed this article then please share it with your friends.



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Force on a Current-Carrying Conductor in Magnetic field – Class 12, JEE & NEET

Force on a Current-Carrying Conductor in Magnetic field

A current-carrying conductor produces a magnetic field around it. i.e. Behaves like a magnet and exerts a force when a magnet is placed in its magnetic field. Similarly a magnet also exerts equal and opposite force on the current-carrying conductor. The direction of this force can be determined using Flemings left-hand rule.

Current Element

Current element is defined as a vector having a magnitude equal to the product of current with a small part of the length of the conductor and the direction in which the current

is flowing in that part of the conductor.

With the help of experiments Ampere established that when a current element 1 de is placed

in a magnetic field $\overrightarrow{\mathrm{B}},$ it experiences a force

$\overrightarrow{\mathrm{dF}}=\mathrm{I} \overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{B}}$

The magnitude of force is $\quad \mathrm{dF}=\mathrm{B} \mathrm{I}$ d\ell $\sin \theta$

$[\theta \text { is the angle between the } \overrightarrow{\mathrm{d} \ell} \text { and } \overrightarrow{\mathrm{B}}]$

(i) When $\sin \theta=\min =0,$ i.e., $\theta=0^{\circ}$ or $180^{\circ}$

Then force on a current element $=0$ (minimum)

A current element in a magnetic field does not experience any force if the current in it is collinear with the field.

(ii) When $\sin \theta=\max =1,$ i.e., $\theta=90^{\circ}$

The force on the current will be maximum ( $=$ BI d\ell)

Force on a current element in a magnetic field is maximum $(=\text { BI } \mathrm{d} \ell)$

When it is perpendicular to the field.

The direction of force is always perpendicular to the plane containing I $\overrightarrow{\mathrm{d} \ell}$ and $\overrightarrow{\mathrm{B}}$.

The direction of the force on current element I $\mathrm{d} \vec{\ell}$ and $\overrightarrow{\mathrm{B}}$ are perpendicular to each other can also be determined by applying either of the following rules:

  1. Fleming’s Left-Hand Rule
  2. Right Hand Palm Rule
Click here for the Video tutorials of Magnetic Effect of Current Class 12

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Solenoid and Toroid | Difference between Solenoid and Toroid
A toroid works as an inductor, which boosts the frequency to appropriate levels. Inductors are electronic components that are passive so that they can store energy in the form of magnetic fields. A toroid turns, and with those turns induces a higher frequency. Toroids are more economical and efficient than solenoids. So here we will study in detail about solenoids and Toroids and the differences between both.
  1. Solenoid
  2. Toroids
  3. Difference between Solenoid and Toroid
  4. Similarities Between Solenoid and Toroid

Toroid

A toroid is an endless solenoid in the form of a ring, as shown in fig.
A toroid is often used to create an almost uniform magnetic field in some enclosed area.
The device consists of a conducting wire wrapped around a ring (a torus) made of a non-conducting material.
Let a toroid having N closely spaced turns of wire,
the magnetic field in the region occupied by the torus = B
Radius of the Toroid ring = R
To calculate the field, we must evaluate $\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} \ell}$ over the circle of radius $\mathrm{R} .$ By symmetry magnitude of the field is constant at the circumference of the circle and tangent to it.

So, $\quad \quad \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d}} \vec{\ell}=\mathrm{B} \ell=\mathrm{B}(2 \pi \mathrm{R})$



This result shows that $\mathrm{B} \propto \frac{1}{\mathrm{R}}$ and hence is nonuniform in the region occupied by torus. However,

if R is very large compared with the cross-sectional radius of the torus, then the field is approximately uniform inside the torus.

Number of turns per unit length of torus $n=\frac{N}{2 \pi R}$

$\therefore \quad B=\mu_{0} n I$

For an ideal toroid, in which turns are closely spaced, the external magnetic field is zero. This is because the net current passing through any circular path laying outside the toroid is zero. Therefore, from Ampere’s law, we find that $B=0,$ in the regions exterior to the torus.

Solenoid:

A long wire wrapped around in a helical shape is known as Solenoid. They are cylindrical in shape as you can see in the image above. A solenoid is used in experiments and research field regarding the magnetic field. The magnetic field inside the solenoid is uniform.

Difference between Solenoid and Toroid

Solenoid and toroid both work on the principle of electromagnetism and both behave like an electromagnet when the current is passed. The magnetic field produces by them is $\mathrm{B}=$ Uo.ni. Even after having so many similarities solenoid and toroid differs in property such as shape:

Solenoid

Toroid

Cylindrical in shape Circular in shape
Magnetic field is created outside Magnetic field is created within
Magnetic field is created outside It does not have a uniform magnetic field inside it.
Has uniform magnetic field inside it. Magnetic feild Outside :- magnetic field (B) $=0$
Magnetic feild due to solenoid is $(B)=$ Uo.nì Magnetic feild Inside:-Magnetic field $(B)=0$
Magnetic feild Within the toroid:- Magnetic field (B) = Uo.ni
 

Similarities between Solenoid and Toroid

  1. Both works on the principle of Electromagnetism.
  2. When the current is passed through them, they both act as an electromagnet.
  3. Magnetic field due to the solenoid and within the toroid is the same. B = Uo.ni


Click here for the Video tutorials of Magnetic Effect of Current Class 12

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Magnetic Field of a Solenoid || Class 12 Physics Notes
The Solenoid is a coil of wire that acts like an electromagnet when a flow of electricity passes through it. Electromagnetic solenoids find use all over the world. You can hardly swing a bat without hitting a solenoid. Speakers and microphones both contain solenoids. In fact, a speaker and microphone are pretty much exactly the same things in reverse of each other. So here we will study in detail about Magnetic Field of a Solenoid Derivation in this article:

Magnetic Field of a Solenoid

A solenoid is a long cylindrical helix. It is made by winding closely a large number of turns of insulated copper wire over a tube of cardboard or china-clay. When an electric current is passed through the solenoid, a magnetic field is produced around and within the solenoid.

The figure shows the lines of force of the magnetic field due to a solenoid. The lines of force inside the solenoid are nearly parallel which indicates that the magnetic field ‘within’ the solenoid is uniform and parallel to the axis of the solenoid.

Magnetic Field of a Solenoid Derivation

Let there be a long solenoid of radius ‘a’ and carrying a current I. Let n be the number of turns per unit length of the solenoid. Let $P$ be a point on the axis of the solenoid.

Let us imagine the solenoid to be divided up into a number of narrow coils and consider one such coil AB of width $\delta x$. The number of turns in this coil is n\deltax. Let $x$ be the distance of the point $P$ from the centre $\mathrm{C}$ of this coil. The magnetic field at $\mathrm{P}$ due to this elementary coil is given by

Let average distance of circumference is r and $\delta \theta$ the angle subtended by the coil at P.

$$ \sin \theta=\frac{\mathrm{BN}}{\mathrm{AB}}=\frac{\mathrm{r} \delta \theta}{\delta \mathrm{x}} \quad \text { or } \quad \delta \mathrm{x}=\frac{\mathrm{r} \delta \theta}{\sin \theta} $$

In $\Delta \mathrm{ACP},$ we have $\quad \mathrm{a}^{2}+\mathrm{x}^{2}=\mathrm{r}^{2}$

$$ r^{3}=\left(a^{2}+x^{2}\right)^{3 / 2} $$

Substiuting these values of $\delta x$ and $\left(a^{2}+x^{2}\right)^{3 / 2}$ in eq. (i), we get

$$ \delta \mathrm{B}=\frac{\mu_{0}(\mathrm{nr} \delta \theta) \mathrm{I} \mathrm{a}^{2}}{2 \mathrm{r}^{3} \sin \theta}=\frac{\mu_{0} \mathrm{nI} \mathrm{a}^{2}}{2 \mathrm{r}^{2}} \frac{\delta \theta}{\sin \theta} $$



The magnetic field $\mathrm{B}$ at $\mathrm{P}$ due to the whole solenoid can be obtained by integrating the above expression between the limits $\theta_{1}$ and $\theta_{2},$ where $\theta_{1}$ and $\theta_{2}$ are the semi-vertical angles subtended at $P$ by the first and the last turn of the solenoid respectively. Thus





or $\quad \mathrm{B}=\frac{1}{2} \mu_{0} \mathrm{n} \mathrm{I}\left(\cos \theta_{1}-\cos \theta_{2}\right)$ ……….(ii)

If the observation point P is well inside a very long solenoid





Thus, the magnetic field at the ends of a ‘long’ solenoid is half of that at the center. If the solenoid is sufficiently long, the field within it (except near the ends) in uniform. It does not depend upon the length and area of the cross-section of the solenoid. Just as a parallel-plate capacitor produces a uniform and known electric field, a solenoid produces a uniform and known magnetic field.

The ‘uniform’ magnetic field within a long solenoid is parallel to the solenoid axis.

Its direction along the axis is given by a curled-straight right-hand rule. “If we grasp the solenoid with our right hand so that our fingers follow the direction of the current in the winding’s, then out extended right thumb will point in the direction of the axial magnetic field”.

  Click here for the Video tutorials of Magnetic Effect of Current Class 12

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