Mind Maps for Rotational Motion Revision – Class XI, JEE, NEET

Class 9-10, JEE & NEET

Rotational Motion Class 11 comprises variety of cases with important formulae and key points. So here is the mind maps to help you in remembering all the key formulae and important concepts on finger tips.

### Mind Map for Moment of Inertia

Download ### Mind Map for Torque

Download ### Mind Maps for Angular Momentum

Download ### Mind Map for Rotational Motion CRTM

Download True Dip and Apparent Dip – Magnetism and Matter Class 12

Class 9-10, JEE & NEET

Here we will study about True Dip and Apparent Dip.

### Apparent Dip

The dip at a place is determined by a dip circle. It consists of magnetized needle capable of rotation in vertical plane about a horizontal axis. The needle moves over a vertical scale graduated in degrees. If the plane of the scale of the dip circle is not in the magnetic meridian then the needle will not indicate correct direction of earth’s magnetic field. The angle made by the needle with the horizontal is called Apparent Dip.

### True Dip

When the plane of scale of dip circle is in the magnetic meridian the needle comes to rest in direction of earth’s magnetic field. The angle made by the needle with the horizontal is called True Dip. Suppose dip circle is set at angle $\alpha$ to magnetic meridian. Horizontal component $${B_H}’ = {B_H}\cos \alpha$$ Vertical component $${B_V}’ = {B_V}$$ (remains unchanged) Apparent dip is $${\theta ^\prime }\tan {\theta ^\prime } = {{{B_V}’} \over {{B_H}’}} = {{{B_V}} \over {{B_H}\cos \alpha }} = {{\tan \theta } \over {\cos \alpha }}\left( {\tan \theta = {{{B_V}} \over {{B_H}}} = {\rm{ true dip }}} \right)$$ 1. For a vertical plane other than magnetic meridian $\alpha>0$ or $\cos \alpha<1$ so $\theta^{\prime}>\theta$ In a vertical plane other than magnetic meridian angle of dip is more than in magnetic meridian.
2. For a plane perpendicular to magnetic meridian $\alpha=\frac{\pi}{2}$ $\therefore \tan \theta^{\prime}=\infty \quad$ so $\quad \theta^{\prime}=\frac{\pi}{2}$ So in a plane perpendicular to magnetic meridian dip needle will become vertical.
At magnetic equator :
1. Angle of dip is zero.
2. Vertical component of earths magnetic field becomes zero $B_{V}=B \sin \theta=B \sin 0=0$
3. A freely suspended magnet will become horizontal at magnetic equator.
4. At equator earth’s magnetic field is parallel to earth’s surface i.e., horizontal.
At magnetic poles :
1. Angle of dip is $90^{\circ}$
2. Horizontal component of earth’s magnetic field becomes zero. $B_{H}=B \cos \theta=B \cos 90=0$
3. A freely suspended magnet will become vertical at magnetic poles.
4. At poles earth’s magnetic field is perpendicular to the surface of earth i.e. vertical.

Ex. If $\theta_{1}$ and $\theta_{2}$ are angles of dip in two vertical planes at right angle to each other and $\theta$ is true dip then prove $\cot ^{2} \theta=\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}$. Sol. If the vertical plane in which dip is $\theta_{1}$ subtends an angle $\alpha$ with meridian than other vertical plane in which dip is $\theta_{2}$ and is perpendicular to first will make an angle of $90-\alpha$ with magnetic meridian. If $\theta_{1}$ and $\theta_{2}$ are apparent dips than $\tan \theta_{1}=\frac{B_{V}}{B_{H} \cos \alpha}$ $\tan \theta_{2}=\frac{B_{V}}{B_{H} \cos (90-\alpha)}=\frac{B_{V}}{B_{H} \sin \alpha}$ $\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\frac{1}{\left(\tan \theta_{1}\right)^{2}}+\frac{1}{\left(\tan \theta_{2}\right)^{2}}=\frac{B_{H}^{2} \cos ^{2} \alpha+B_{H}^{2} \sin ^{2} \alpha}{B_{V}^{2}}=\frac{B_{H}^{2}}{B_{V}^{2}}=\left(\frac{B \cos \theta}{B \sin \theta}\right)^{2}=\cot ^{2} \theta$ So $\quad \cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\cot ^{2} \theta$
Also Read: Properties of Paramagnetic & Diamagnetic Materials
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Neutral Point of Magnet – Magnetism and Matter Class 12 Physics Notes

Class 9-10, JEE & NEET

A neutral point of Magnet is a point at which the resultant magnetic field is zero. In general the neutral point is obtained when horizontal component of earth’s field is balanced by the produced by the magnet. When the N pole of the magnet points South and the magnet in the magnetic meridian. When we plot magnetic field of a bar magnet the curves obtained represent the superposition of magnetic fields due to bar magnet and earth. DEFINITION A neutral point in the magnetic field of a bar magnet is that point where the field due to the magnet is completely neutralized by the horizontal component of earth’s magnetic field. At neutral point field due to bar magnet (B) is equal and opposite to horizontal component of earth’s magnetic field $\left( B _{ H }\right) or \quad B = B _{ H }$
• #### Neutral point when north pole of magnet is towards geographical north of earth. The neutral points $N_{1}$ and $N_{2}$ lie on the equatorial line. The magnetic field due to magnet at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$ Where M is magnetic dipole moment of magnet, $2 l$is its length and r is distance of neutral point.

At neutral point $B = B _{ H } \cdot \quad$ so $\quad \frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}= B _{ H }$

• #### Neutral point when south pole of magnet is towards geographical north of earth. The neutral points $N_{1}$ and $N_{2}$ lie on the axial line of magnet. The magnetic field due to magnet

at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$

At neutral point $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$ so $\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}= B _{ H }$

#### Special Point

When a magnet is placed with its S pole towards north of earth neutral points lie on its axial line. If magnet is placed with its N pole towards north of earth neutral points lie on its equatorial line. So neutral points are displaced by $90^{\circ}$ on rotating magnet through $180^{\circ}$ In general if magnet is rotated by angle $\theta$ neutral point turn through an angle $\frac{\theta}{2}$

#### Neutral Point in Special Cases (a) If two bar magnets are placed with their axis parallel to each other and their opposite poles face each other then there is only one neutral point (x) on the perpendicular bisector of the axis equidistant from the two magnets.

(b) If two bar magnets are placed with their axis parallel to each other and their like poles face each other then there are two neutral points on a line equidistant from the axis of the magnets.

Also Read: Properties of Paramagnetic & Diamagnetic Materials
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Bar Magnet as an Equivalent Solenoid – Magnetism || Class 12 Physics Notes

Class 9-10, JEE & NEET

###  In solenoid each turn behaves as a small magnetic dipole having dipole moment $\mathrm{I} \mathrm{A}$. A solenoid is treated as arrangement of small magnetic dipoles placed in line with each other. The number of dipoles is equal to number of turns in a solenoid. The south and north poles of each turn cancel each other except the ends. So solenoid can be replaced by single south and north pole separated by distance equal to length of solenoid. The magnetic field produced by a bar magnet is identical to that produced by a current carrying solenoid. Derivation of Bar Magnet as an Equivalent Solenoid To demonstrate the similarity of a current carrying solenoid to a bar magnet, let us calculate axial field of a finite solenoid carrying current. Consider: $a=$ radius of solenoid

$2 l=$ length of solenoid with centre O

$n=$ number of turns per unit length $I=$ current passing through solenoid

$O P=r$

Consider a small element of thickness $d x$ of solenoid at distance $x$ from O. and number of turns in element $=n d x$

We know magnetic field due to n turns coil at axis of solenoid is given by

$d B=\frac{\mu_{0} n d x I a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{\frac{3}{2}}}$

The magnitude of the total field is obtained by summing over all the elements $-$ in other words by integrating from $x=-1$ to $x=+1 .$ Thus,

$B=\frac{\mu_{0} n I a^{2}}{2} \int_{-1}^{l} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}$

This integration can be done by trigonometric substitutions. This exercise, however, is not necessary for our purpose. Note that the range of $x$ is from $-1$ to $+1 .$ Consider the far axial field of the solenoid, i.e., $r>>$ a and $r>>1 .$ Then the denominator is approximated by

\begin{aligned}\left[(r-x)^{2}+a^{2}\right]^{3 / 2} &=r^{3} \\ \text { and } B &=\frac{\mu_{0} n I a^{2}}{2 r^{3}} \int_{-1}^{1} d x \\ &=\frac{\mu_{0} n I}{2} \frac{2 l a^{2}}{r^{3}} \end{aligned}

Note that the magnitude of the magnetic moment of the solenoid is, (total number of turns $\times$ current $\times$ cross-sectional area). Thus,

$B=\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}$

It is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Terrestrial Magnetism – Earth’s Magnetism || Class 12 Physics Notes

Class 9-10, JEE & NEET

Do you know that the Earth is also a magnet? Yes!!! How do you think then that the suspended bar magnet always points in the north-south direction? Adware about the concept of the Terrestrial Magnetism that we are going to discuss in this chapter. It is really interesting to study and analyze this concept of earth’s magnetism. The branch of Physics which deals with the study of earth’s magnetic field is called terrestrial magnetism.
1. William Gilbert suggested that earth itself behaves like a huge magnet. This magnet is so oriented that its S pole is towards geographic north and N pole is towards the geographic south.
2. The earth behaves as a magnetic dipole inclined at small angle $11.5^{\circ}$ to the earth’s axis of rotation with its south pole pointing geographic north.
3. The idea of earth having magnetism is supported by following facts.
4. A freely suspended magnet always comes to rest in N-S direction.
5. A piece of soft iron buried in N-S direction inside the earth acquires magnetism.
6. Existence of neutral points. When we draw field lines of bar magnet we get neutral points where magnetic field due to magnet is neutralized by earth’s magnetic field.
7. The magnetic field at the surface of earth ranges from nearly 30 $\mu T$ near equator to about 60$\mu T$ near the poles. The magnetic field on the axis is nearly twice the magnetic field on the equatorial line.
Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Magnetic Elements – Magnetism Class 12 || Physics Notes

Class 9-10, JEE & NEET

The physical quantities which determine the intensity of earth’s total magnetic field completely both in magnitude and direction are called magnetic elements.

### Angle of Declination $(\phi)$:

The angle between the magnetic meridian and geographic meridian at a place is called angle of declination.

(a) Isogonic Lines : Lines drawn on a map through places that have same declination are called isogonic lines.

(b) Agonic Line : The line drawn on a map through places that have zero declination is known as an agonic line.

###  Angle of Dip or Inclination

The angle through which the N pole dips down with reference to horizontal is called the angle of dip. At magnetic north and south pole angle of dip is $90^{\circ}$. At magnetic equator the angle of dip is zero.

OR The angle which the direction of resultant field of earth makes with the horizontal line of magnetic meridian is called angle of dip. (a) Isoclinic Lines : Lines drawn up on a map through the places that have same dip are called isoclinic lines.

(b) Aclinic Line : The line drawn through places that have zero dip is known as aclinic line. This is the magnetic equator.

### Horizontal component of earth’s magnetic field

The total intensity of the earth’s magnetic field makes an angle  with horizontal. It has

(i) Component in horizontal plane called horizontal component $B_{H}$.

(ii) Component in vertical plane called vertical component $B_{V}$

$B _{ V }= B \sin \theta \quad B _{ H }= B \cos \theta$

So $\frac{ B _{ V }}{ B _{ H }}=\tan \theta \quad$ and $\quad B =\sqrt{ B _{ H ^{2}}+ B _{ V ^{2}}}$

#### IMPORTANT POINTS

1. If $\theta$ and $\phi$ are known we can find direction of B.
2. If $\theta$ and $B_{H}$ are known we can find magnitude of B.
3. So if $\theta$, $\phi$ and $B_{H}$ are known we can find total field at a place. So these are called as Elements of earth’s magnetic field.
4. The declination gives the plane, dip gives the direction and horizontal component gives magnitude of earth’s magnetic field.
5. If declination is ignored, then the horizontal component of earth’s magnetic field is from geogrophic south to geographic north.
6. Angle of dip is measured by instrument called dip circle.
Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Bohr Magneton: Unit of Bohr Magneton – Magnetism || Class 12 Physics Notes

Class 9-10, JEE & NEET

Bohr Magneton is defined as the magnetic dipole moment associated with an atom due to orbital motion of an electron in the first orbit of hydrogen atom. This is the smallest value of magnetic moment. Unit of Bohr Magneton is-
• In CGS units it is defined by the equation:

$\mu \mathrm{B}=(e \cdot h) / 2 m_{\mathrm{e}} \mathrm{c}$

• In SI units it is defined by the equation:

$\mu_{\mathrm{B}}=e \cdot h / 2 m_{e}$

1. The electron possesses magnetic moment due to its spin motion also $\overrightarrow{ M }_{ s }=\frac{ e }{ m _{ e }} \overrightarrow{ s }$ .where $\overrightarrow{ s }$ is spin angular momentum of electron and $S=\pm \frac{1}{2}\left(\frac{h}{2 \pi}\right)$
2. The total magnetic moment of electron is the vector sum of its magnetic moments due to orbital and spin motion.
3. The resultant angular momentum of the atom is given by vector sum of orbital and spin angular momentum due to all electrons. Total angular momentum $\vec{J}=\vec{L}+\vec{S}$
4. The resultant magnetic moment $\overrightarrow{ M _{j}}=- g \left(\frac{ e }{2 m }\right) \overrightarrow{ J }$

where g is Lande’s splitting factor which depends on state of an atom.

For pure orbital motion g = 1 and pure spin motion g = 2.

Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Current loop as a Magnetic Dipole – Magnetism | Class 12th Physics Notes

Class 9-10, JEE & NEET

#### 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 show 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 is 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 anti clockwise direction. 1. The magnetic field lines due to each elementary portion of the circular coil are circular near the element and almost straight near center of circular coil. 2. The magnetic lines of force seem to enter at lower face of coil and leave at upper face.
3. The lower face through which lines of force enter behaves as south pole and upper face through which field lines leave behaves as 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. Magnetic dipole moment is a vector with direction from S pole to N pole or along direction of normal to 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 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 $\mu_{ B }=\frac{ eh }{4 \pi m _{ e }}=\frac{\left(1.6 \times 10^{-19} C \right)\left(6.62 \times 10^{-34} Js \right)}{4 \times 3.14 \times\left(9.1 \times 10^{-31} kg \right)}=9.27 \times 10^{-24} Am ^{2}$ is called Bohr Magneton. This is natural unit of magnetic moment.     Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Class 12 Magnetism – Gauss Law Definition || Solved Examples

Class 9-10, JEE & NEET

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

Class 9-10, JEE & NEET

Potential Energy of magnetic dipole in 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 direction of field. If 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 $W =\int d W =\int_{\theta_{1}}^{\theta_{2}} \tau d \theta= MB \int_{\theta_{1}}^{\theta_{2}} \sin \theta d \theta= MB \left(\cos \theta_{1}-\cos \theta_{2}\right)$ 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 field to any given direction. $U = W _{ \theta }- W _{\frac{\pi}{2}}=- MB \cos \theta=-\overrightarrow{ M } \cdot \overrightarrow{ B }$ Also Read: Biot Savart’s Law
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Magnetic Field due to a Short Bar Magnet | Magnetic Dipole – Class 12 physics Notes

Class 9-10, JEE & NEET

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.

## 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. $\overrightarrow{ B }_{\text {axial }}=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$ $\overrightarrow{ B }_{1}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r -\ell)^{2}}(\hat{ r })$ and $\overrightarrow{ B }_{2}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r +\ell)^{2}}(-\hat{ r })$ $\therefore \quad \overrightarrow{ B }_{ axial }=\left[\frac{\mu_{0}}{4 \pi} \frac{ m }{( r -\ell)^{2}}-\frac{\mu_{0}}{4 \pi} \frac{ m }{( r +\ell)^{2}}\right] \hat{ r }=\frac{\mu_{0} m }{4 \pi}\left[\frac{4 r \ell}{( r -\ell)^{2}( r +\ell)^{2}}\right] \hat{ r }$ $\overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ Mr }}{\left( r ^{2}-\ell^{2}\right)^{2}}$ 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 $\ell<< r \overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ M }}{r^{3}}$

## 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 $\overrightarrow{ B }_{ equi }=\overrightarrow{ B _{1}} \cos \theta(-\hat{ r })+\overrightarrow{ B _{2}} \cos \theta(-\hat{ r })=\left[\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cos \theta+\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cos \theta\right](-\hat{ r })$ $=2 \cdot \frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cdot \frac{\ell}{\sqrt{ r ^{2}+\ell^{2}}}(-\hat{ r })$ $\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$ $\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$ 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 $\theta^{\prime}=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)$ Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Coulomb’s law of Magnetism || Magnetism and Matter Class 12, JEE & NEET

Class 9-10, 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.

### Coulomb’s Law in Magnetism If two magnetic poles of strengths $${{m_1}}$$ and $${{m_2}}$$ are kept at a distance r apart then 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 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 north pole is defined as the force experienced by the pole when kept in 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 nature of material of magnet, state of magnetisation (with an upper limit called saturation) and area of cross-section. (4) The north pole experiences a force in the direction of magnetic field while south pole experiences force opposite to field.
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ATOMIC THEORY OF MAGNETISM || Magnetism and Matter Class 12, JEE & NEET

Class 9-10, 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 behaves as 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. 1. In magnetised substance all the atomic magnets are aligned in 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 than each part behaves like a complete magnet. (3) Magnetisation of an electromagnet can be explained as alignment of atomic magnets in direction of 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

Class 9-10, 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 magnet is suspended its length becomes parallel to 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 magnetic axis and the vertical plane passing through the axis of a freely suspended or pivoted magnet is called 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 then 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 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 magnetised 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 magnetisation of a bar, temporarily identical poles at the two ends with an opposite pole of double strength at the centre of 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 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 unmagnetised magnetic material due to ‘induction effect’. (8) Magnetic Induction : A magnet attracts certain other substances through the phenomenon of magnetic induction i.e., by inducing opposite pole in a 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, aluminium 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 attracing 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) Demagnetisation : A magnet gets demagnetised, 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-demagnetisation due to poles at the ends which tends to neutralise each other. However, by using pieces of soft iron called keepers, the poles at the ends are neutralised and consequently the demagnetising effect disappears and the magnet can retain its magnetism for a longer period. Construction and Working Principle of Cyclotron Class 12 Physics

Class 9-10, JEE & NEET

We will study here about the Construction and Working Principle of Cyclotron Class 12

### Cyclotron Introduction

A cyclotron is used for accelerating positive ions, so that they acquire energy large enough to carry out nuclear reactions. 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 spiral path 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 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 ionisation of the gas. To produce proton, hydrogen gas is used; while for producing $\alpha$ -particles, helium gas is used. Theory : Consider that a positive ion is produced at the centre of the gap at the time, when the dee $D_{1}$ is at 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 magnetic field provides the centripetal force to the positive ion and it is deflected along a circular path because 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 in correct phase.

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

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Force Between Two Parallel Current Carrying Conductors || Class 12 Physics Notes

Class 9-10, JEE & NEET

It is experimentally established fact that two current carrying conductors attract each other when the current is in same direction and repel each other when the current 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 $B_{1}=\frac{\mu_{0} I_{1}}{2 \pi d}$ …….(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 The vector product $\overrightarrow{\mathrm{d}} \ell \times \overrightarrow{\mathrm{B}}$ has a direction towards the wire $W_{1}$ . 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 $\operatorname{eq}^{n}(i i)$ If the parallel wires currents in opposite directions, the wires repel each other. The wires attract each other if current in the wires are flowing in the same direction. And they repel each other if the currents are in opposite directions.

#### Experimental Demonstration ### 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   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Force on a Moving Charge in a Magnetic Field | Class 12 Physics Notes

Class 9-10, JEE & NEET

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 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 $\mathrm{d} \tau=\mathrm{S} \mathrm{d} \ell]$ $n \mathrm{d} \tau=$ 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 $180^{\circ}$ (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 magnitude of charge with velocity ### Difference in Force on a Charged Particle by Magnetic Field and Electric Field Also Read: Biot Savart’s Law   Click here for the Video tutorials of Magnetic Effect of Current Class 12
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Motion of Charged Particle in a Magnetic Field | Moving Charges and Magnetism Class 12, JEE & NEET

Class 9-10, 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. Motion of a charged particle in magnetic field is characterized by the change in the direction of motion. It is expected also as magnetic field is capable of only changing direction of motion. In order to keep the context of study simplified, we assume magnetic field to be uniform. This assumption greatly simplifies the description and lets us easily visualize the motion of a charged particle in magnetic field. Motion of a Charged Particle in a Magnetic Field Motion of a charged particle when it is moving collinear with the field magnetic field is not affected by the field (i.e. if motion is just along or opposite to magnetic field) ( : $: \quad F=0$ ) Only the following two cases are possible:   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,  n case of circular motion of a charged particle in a steady magnetic field : i.e., with 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. motion of charged particle in 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)   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   Click here for the Video tutorials of Magnetic Effect of Current Class 12
About eSaral At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.
Flemings Left Hand Rule Class 12 Physics

Class 9-10, JEE & NEET

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 middle finger points in the direction of magnetic field then the thumb points towards the direction of the force or motion of the conductor. Then if the fore-finger 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 force. (ii) Right-hand Palm Rule : Stretch the fingers and thumb of 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. 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 $\vec{F}=I\left[\int d \vec{\ell}\right] \times \vec{B}=I \ell^{\prime} \times \vec{B}$ 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 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
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Right Hand Palm Rule – Magnetic Effect of Current Class12, JEE & NEET

So far we have described the magnitude of the magnetic force on a moving electric charge, but not the direction. The magnetic field is a vector field, thus the force applied will be oriented in a particular direction. There is a clever way to determine this direction using nothing more than your right hand. The direction of the magnetic force $F$ is perpendicular to the plane formed by $v$ and $B$, as determined by the right hand palm rule, which is illustrated in the figure. The right hand rule states that ito determine the direction of the magnetic force on a positive moving charge, $f$, point the thumb of the right hand in the direction of $v,$ the fingers in the direction of $B,$ and a perpendicular to the palm points in the direction of $F$
If we hold the thumb of right hand mutually perpendicular to the grip of the fingers such that the curvature of the finger represents the direction of current in the wire loop, then the thumb of the right hand will point in the direction of magnetic field near the centre of the current loop Graph of B v/s x As soon as x increases magnetic field B decreases, dependence of B on x is shown in figure Rate of change of B with respect to x is different at different values of x for $x<\pm \frac{a}{2}$ curve is convex and $\quad$ for $x>\pm \frac{a}{2}$ curve is concave At $x=\pm \frac{a}{2} \quad$ we get $\frac{d B}{d x}=$ const $,$ and $\frac{d^{2} B}{d x^{2}}=0$ So at $x=+\frac{a}{2} \&-\frac{a}{2}$ B varies linearly with $x$ These points are called point of inflexion. Distance in between these two points is equal to radius of the coil $B=\frac{B_{C}}{\left(1+\frac{x^{2}}{a^{2}}\right)^{3 / 2}}$ $\because$ Magnetic field at the centre of coil $\mathrm{B}_{\mathrm{C}}=\frac{\mu_{0} \mathrm{NI}}{2 \mathrm{a}}$ 