Here is the Mind Maps for Class 12Th Electrostatics Comprising all the important Formulae and Key Points. Therea re very important for quick revision during Exam.Mind Map of Electric Field and Charges
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Mind Map of Electric Potential
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Ray Optics in Class 12th comprises variety of cases with important formula and key points. So here is the mind map to help you in remembering all the formula and important key concepts on finger tips.
Mind Map of Reflection of Light
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DownloadAtomic Structure in Class 12 comprises variety of cases with important formulae and key points. So here are the mind maps to help you in remembering all the formulas and important key concepts on finger tips.
Mind Map of Atomic Structure
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MIND MAP 1 Download MIND MAP 2 DownloadMind Map of Radioactivity
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DownloadHere 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 $${\theta ^\prime }\tan {\theta ^\prime } = {{{B_V}’} \over {{B_H}’}} = {{{B_V}} \over {{B_H}\cos \alpha }} = {{\tan \theta } \over {\cos \alpha }}\left( {\tan \theta = {{{B_V}} \over {{B_H}}} = {\rm{ true dip }}} \right)$$ 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.
 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.
 Angle of dip is zero.
 Vertical component of earths magnetic field becomes zero $B_{V}=B \sin \theta=B \sin 0=0$
 A freely suspended magnet will become horizontal at magnetic equator.
 At equator earth’s magnetic field is parallel to earth’s surface i.e., horizontal.
 Angle of dip is $90^{\circ}$
 Horizontal component of earth’s magnetic field becomes zero. $B_{H}=B \cos \theta=B \cos 90=0$
 A freely suspended magnet will become vertical at magnetic poles.
 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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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
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
$d B=\frac{\mu_{0} n d x I a^{2}}{2\left[(rx)^{2}+a^{2}\right]^{\frac{3}{2}}}$
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[(rx)^{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[(rx)^{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$ crosssectional 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 12About eSaral At eSaral we are offering a complete platform for IITJEE & 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.For free video lectures and complete study material, Download eSaral APP.
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 northsouth 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.
 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.
 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.
 The idea of earth having magnetism is supported by following facts.
 A freely suspended magnet always comes to rest in NS direction.
 A piece of soft iron buried in NS direction inside the earth acquires magnetism.
 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.
 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.
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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
 If $\theta$ and $\phi$ are known we can find direction of B.
 If $\theta$ and $B_{H}$ are known we can find magnitude of B.
 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.
 The declination gives the plane, dip gives the direction and horizontal component gives magnitude of earth’s magnetic field.
 If declination is ignored, then the horizontal component of earth’s magnetic field is from geogrophic south to geographic north.
 Angle of dip is measured by instrument called dip circle.
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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}$
 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)$
 The total magnetic moment of electron is the vector sum of its magnetic moments due to orbital and spin motion.
 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}$
 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 12About eSaral At eSaral we are offering a complete platform for IITJEE & 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.For free video lectures and complete study material, Download eSaral APP.
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. 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.
 The magnetic lines of force seem to enter at lower face of coil and leave at upper face.
 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.
 A planar loop of any shape behaves as a magnetic dipole.
 The dipole moment of current loop $(\mathrm{M})=$ ampere turns (nI) $\times$ area of coil (A) or $\mathrm{M}=\mathrm{nIA}$.
 The unit of dipole moment is ampere meter $^{2}\left( A – m ^{2}\right)$
 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 12About eSaral At eSaral we are offering a complete platform for IITJEE & 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.For free video lectures and complete study material, Download eSaral APP.
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$
 Isolated magnetic poles do not exist is a direct consequence of gauss law in magnetism.
 The total magnetic flux linked with a closed surface is always zero.
 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 midpoint. 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 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 LawAbout eSaral At eSaral we are offering a complete platform for IITJEE & 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.For free video lectures and complete study material, Download eSaral APP.
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 Npole of magnet and $\overrightarrow{ B _{2}}$ due to Spole 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}}$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 Npole of magnet and $\overrightarrow{ B _{2}}$ due to Spole of magnet $\overrightarrow{ B }_{ equi }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$ due to Spole 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)$Also Read:
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Magnetic Field is defined as 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 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 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 same at all the points in that region. (3) The strength of magnetic field is also known as magnetic induction or magnetic flux density. (4) The $\mathrm{SI}$ unit of strength of 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)$
 The CGS unit is Gauss (G)
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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 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 magnetic axis. (3) The distance between the two poles of a bar magnet is called the magnetic length of magnet. It is denoted by 2$\ell$ (4) The distance between the ends of the magnet is called 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 magnetic dipole moment. $\overrightarrow{ M }= m (\overrightarrow{2 \ell})$ Important Points to Remember (1) It is a vector quantity whose direction is from south pole to north pole of 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 original value. 4. If a magnet is cut into two equal parts transverse to the length then pole strength remains unchanged and length is reduced to half. 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 original value. (5) In magnetism existence of magnetic monopole is not possible. (6) The magnetic dipole moment of a magnet is equal to product of pole strength and distance between poles. M = m d (7) As magnetic moment is a vector, in 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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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 (Am) 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. $m =\frac{\overrightarrow{ F }}{\overrightarrow{ B }}$ Pole strength is a scalar quantity with dimension $M^{0} L^{1} T^{0} A^{1}$
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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 :
 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.
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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 (Fe_{3}O_{4}), 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 horseshoe 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 :
 The permanent artificial magnets are made of some metals and alloys like carbonsteel, Alnico, Platinumcobalt, Alcomax, Ticonal. The permanent magnets are made of ferromagnetic substances with large coercivity and retentivity and can have desired shape like barmagnet, U shaped magnet or magnetic needle etc. These magnet retain its attracting power for a long time.
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Here are some of the Limitations of Cyclotron listed below:
 When a charged particle is accelerated, its mass also starts increasing with increase in its speed. When its speed become comparable to that of light, the mass of the charged particle become quite large as compared to its rest mass.
If $\mathrm{m}_{0}$ = rest mass, mass of charged particle moving with speed $v$ is $\quad m=\frac{m_{0}}{\sqrt{1\frac{v^{2}}{c^{2}}}}$
Substituting for $\mathrm{m}$ in equation, we have
time spent inside a dee, $t=\frac{\pi \mathrm{m}_{0}}{\mathrm{qB} \sqrt{1\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$
Therefore, as v increases, t also increases i.e. the charged particle starts taking more and more time to complete the semicircular path inside the dee.
Since electric field changes the polarity of the dees after a fixed interval, the charged particle starts lagging behind the electric field and it us ultimately lost by colliding against the walls of the dees.\
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However, this problem is overcome in the following two ways :
(a) As $v$ increases, $\sqrt{1\frac{v^{2}}{c^{2}}}$ decreases. Therefore, $B$ is increased in such a manner that the facror $\mathrm{B} \sqrt{1\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}$ and hence t always remains constatn. Such a cyclotron, in which the strength of magnetic field is adjusted to eovercome the problem due to relative variation in mass of the positive ion, is called synchrotron.
(b) The frequency of revolution of charged [particle inside the dees] may be expressed as
$v=B q \sqrt{\frac{1\frac{v^{2}}{c^{2}}}{2 \pi m_{0}}}$
It follows that as $v$ increases, $\sqrt{1\frac{v^{2}}{c^{2}}}$ decreases and hence v decreases. If frequency of the electric field is adjusted to be always equal to the frequency of revolution of the charged particle, then such a cyclotron is called synchrocyclotron or frequency modulated cyclotron.
2. Cyclotron is used to accelerate heavy charged particles, such as protons. It is not suitable for accelerating electrons.
The reason is that due to small mass, the electrons gain in speed quickly and likewise the relativistic variation in mass quickly makes them out of step with the oscillating electric field.
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Also Read
 Right Hand Thumb Rule
 Biot Savart’s Law
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 Helmholtz Coils  Magnetic Field between two Coils
 Magnetic Field at the Centre of a Circular Coil
 Ampere’s Circuital Law and it Applications
 Magnetic Field of a Solenoid  Class 12 Physics Notes
 Solenoid and Toroid  Difference between Solenoid and Toroid
 Force on a Current Carrying Conductor in a Magnetic Field
 Flemings Left Hand Rule Class 12 Physics
 Torque on a Current loop in a Magnetic Field
 Motion of Charged Particle in a Magnetic Field
 Motion of Charged Particle in Electric and Magnetic Field
 Force on a Moving Charge in a Magnetic Field
 Force Between Two Parallel Current Carrying Conductors
 Cyclotron Working Principle, Applications & Frequency
 Construction and Working Principle of Cyclotron
 Limitations of Cyclotron
 Magnetic Field due to Infinite Straight Conductor