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The Dipole In A Uniform Magnetic Field - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

Dipole in a uniform magnetic field is considered one the most difficult concept.
21 Questions around this concept.

Solve by difficulty

EASY

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60⁰. The torque needed to maintain the needle in this position will be

$\sqrt{3}W$

$W$

$\left ( \frac{\sqrt{3}}{2} \right )W$

$2W$

View Solution

Concepts Covered - 1

Dipole in a uniform magnetic field

Net Force-

As magnetic dipole is analogous to an electric dipole.

So we can use $m=q_m$

when a magnetic dipole is kept in a uniform magnetic field. The net force experienced by the dipole is zero as shown in the below figure.

I.e $F_{net}=0$

Hence magnetic dipole will not make any linear motion.

Torque on dipole-

Net torque about the center of dipole is given as $\tau =q_mB(2a)\sin\theta$

Using $\vec{M}=q_m2a$ we get $\tau =MB\sin\theta$

So $\vec{\tau }=\vec{M}\times \vec{B}$

The direction of the torque is normal to the plane containing dipole moment M and magnetic field B and is governed by right-hand screw rule.
If Dipole is parallel to B the torque is Zero. I.e $\Theta =0^{\circ}$ $\tau =0$ (This is the position of stable equilibrium of dipole)
Torque is maximum when Dipole is perpendicular to B. I.e $\Theta =\frac{\pi }{2}$ $\tau =MB= maximum \ \ \ torque$

Oscillation of dipole -If a dipole experiencing a torque in a magnetic field is allowed to rotate, then it will rotate to align itself to the magnetic field. But when it reaches along the direction of B the torque becomes zero. But due to inertia, it overshoots this equilibrium condition and then starts oscillating about this mean position.

The time period of this oscillation is given as

$T=2\pi \sqrt{\frac{I}{MB}}$

where I= moment of inertia of dipole about the axis passing through its center and perpendicular to its length.

For two magnets having Magnetic Moments in the same direction (i.e sum position of the magnetic moment)

$M_{s}=M_{1}+M_{2}$

$I_{s}=I_{1}+I_{2}$

$M_{s}$ - Net Magnetic Moment

$I_{s}$ - Net Moment of Inertia

So Time period is

$\dpi{100} T=2\pi \sqrt{\frac{I_s}{M_sB}}$

Similarly, Frequency is given as

$\nu =\frac {I}{T_{s}}=\frac{1}{2\pi}\sqrt \frac{(M_{1}+M_{2})B}{I_{s}}$

For two magnets having Magnetic Moments in the opposite direction (i.e difference position of the magnetic moment)

$M_{d}=M_{1}-M_{2}$

$I_{d}=I_{1}+I_{2}$

So Time period is

$T=2\pi \sqrt{\frac{I_d}{M_dB}}$ or $T_{d}=2\pi\sqrt \frac{I_{1}+I_{2}}{(M_{1}-M_{2})B}$

Similarly, Frequency is given as

$\nu_{d}=\frac{1}{T_{d}}=\frac{1}{2\pi}\:\sqrt \frac{(M_{1}-M_{2})B}{I_{1}+I_{2}}$

The ratio of difference and sum position of the magnetic moment

$\frac{T_{s}}{T_{d}}=\sqrt{\frac{M_{1}-M_{2}}{M_{1}+M_{2}}}$

$\frac{M_{1}}{M_{2}}=\frac{T_{d}^{2}+T_{s}^{2}}{T_{d}^{2}-T_{s}^{2}}=\frac{\nu _{s}^{2}+ \nu _{d}^{2}}{\nu _{s}^{2}-\nu _{d}^{2}}$

Dipole in Non-Uniform magnetic field- In case the magnetic field is non-uniform, the magnitude of the force on + $+q_m \ \ \ and \ \ \ - q_m$ will be different. So $F_(net)\neq0$ and At the same time due to a couple of forces acting, a torque will also be acting on it.

Work done in rotation-

Then work done by magnetic force for rotating a magnetic dipole through an angle $\theta _2$ from the equilibrium position of an angle $\theta _1$ (As shown in the above figure) is given as

$\\ W_{mag}=\int \tau d\theta =\int_{\theta _1}^{\theta _2}\tau d\theta cos(180^0)=-\int_{\theta _1}^{\theta _2}\tau d\theta \\ \Rightarrow W_{mag}=-\int_{\theta _1}^{\theta _2}(M\times B) d\theta=-\int_{\theta _1}^{\theta _2}(MBSin\theta ) d\theta=MB\left ( \cos \Theta _{2} -\cos \Theta _{1}\right )$

And So work done by an external force is $W=-W_{mag}=MB\left ( \cos \Theta _{1} -\cos \Theta _{2}\right )$

For example

if $\Theta _{1}=0^{\circ} \; and\; \Theta _{2}=\Theta$

$W=MB \left ( 1-\cos \Theta \right )$

$if\; \Theta _{1}= 90^{\circ}\; and \; \Theta _{2}=\Theta$

$W=-MB\cos \Theta$

Potential Energy of a dipole kept in a magnetic field-

As $\Delta U=-W_{mag}=W$

So change in Potential Energy of a dipole when it is rotated through an angle $\theta _2$ from the equilibrium position of an angle $\theta _1$ is given as $\Delta U=MB\left ( \cos \Theta _{1} -\cos \Theta _{2}\right )$