 Rotational Motion: Moment of Inertia
 Rotational Motion: Torque
 Rotational Motion: Angular Momentum
 Rotational Motion: CRTM
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
About 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.

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
About 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.
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.
 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.
About 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.
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.
About 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.
 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.
 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
About 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.
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.
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 $$ Magnetic flux density $B _{2}=\frac{\mu_{0}}{4 \pi} \frac{ M \sin \theta}{ r ^{3}}$ Total magnetic flux density at P. $B =\sqrt{ B _{1}^{2}+ B _{2}^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{4 \cos ^{2} \theta+\sin ^{2} \theta}$ or $\tan \theta^{\prime}=\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{r^{3}}}{\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}}}=\frac{1}{2} \tan \theta$ or $\theta^{\prime}=\tan ^{1}\left(\frac{1}{2} \tan \theta\right)$Also Read:
Click here for the Video tutorials of Magnetic Effect of Current Class 12About eSaral At eSaral we are offering a complete platform for IITJEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.For free video lectures and complete study material, Download eSaral APP.
Coulomb’s Law in Magnetism
If two magnetic poles of strengths $${{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 $M^{0} L^{1} T^{0} A^{1}$
About 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.
 In magnetised substance all the atomic magnets are aligned in same direction and thus resultant magnetism is nonzero.
About 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.
 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.
For free video lectures and complete study material, Download eSaral APP.
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 Dshaped 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 semicircular 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 semicircular 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 semicircular 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 semicircular 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.Also Read:
Click here for the Video tutorials of Magnetic Effect of Current Class 12
About 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.
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 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.
 Force on a Charged Particle in a Magnetic Field
 Difference in Force on a Charged Particle by Magnetic Field and Electric Field
Difference in Force on a Charged Particle by Magnetic Field and Electric Field
Also Read: Biot Savart’s Law 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.
About 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 in a Uniform Field
(e) if a currentcarrying conductor is situated in a nonuniform 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 zeroClick here for the Video tutorials of Magnetic Effect of Current Class 12
About 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.