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

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.
।.e $F_{n e t}=0$

 

Hence magnetic dipole will not make any linear motion.

Torque on dipole-

Net torque about the center of dipole is given as $\tau=q_m B(2 a) \sin \theta$
Using $\vec{M}=q_m 2 a_{\text {we get }} \tau=M B \sin \theta$
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 $\Theta=0^{\circ} \quad \tau=0$ (This is the position of stable equilibrium of dipole)
- Torque is maximum when Dipole is perpendicular to B. I.e $\Theta=\frac{\pi}{2} \quad \tau=M B=$ maximum torque starts oscillating about this mean position.

The time period of this oscillation is given as

$
T=2 \pi \sqrt{\frac{I}{M B}}
$

where $\mathrm{I}=$ 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)

$
\begin{aligned}
& M_s=M_1+M_2 \\
& I_s=I_1+I_2
\end{aligned}
$

$M_s$ - Net Magnetic Moment
$I_s$ - Net Moment of Inertia
So Time period is

$
T=2 \pi \sqrt{\frac{I_s}{M_s B}}
$

Similarly, Frequency is given as

$
\nu=\frac{I}{T_s}=\frac{1}{2 \pi} \sqrt{\frac{\left(M_1+M_2\right) B}{I_s}}
$

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

$
\begin{aligned}
& M_d=M_1-M_2 \\
& I_d=I_1+I_2
\end{aligned}
$
 

So Time period is 

$
T=2 \pi \sqrt{\frac{I_d}{M_d B}} \quad T_d=2 \pi \sqrt{\frac{I_1+I_2}{\left(M_1-M_2\right) B}}
$

Similarly, Frequency is given as

$
\nu_d=\frac{1}{T_d}=\frac{1}{2 \pi} \sqrt{\frac{\left(M_1-M_2\right) B}{I_1+I_2}}
$

- The ratio of difference and sum position of the magnetic moment

$
\begin{aligned}
& \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}
\end{aligned}
$
 

  

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

$
\begin{aligned}
& W_{m a g}=\int \tau d \theta=\int_{\theta_1}^{\theta_2} \tau d \theta \cos \left(180^0\right)=-\int_{\theta_1}^{\theta_2} \tau d \theta \\
& \Rightarrow W_{\text {mag }}=-\int_{\theta_1}^{\theta_2}(M \times B) d \theta=-\int_{\theta_1}^{\theta_2}(M B \operatorname{Sin} \theta) d \theta=M B\left(\cos \Theta_2-\cos \Theta_1\right)
\end{aligned}
$

And So work done by an external force is $W=-W_{m a g}=M B\left(\cos \Theta_1-\cos \Theta_2\right)$
For example

$
\begin{aligned}
& \text { if } \Theta_1=0^{\circ} \text { and } \Theta_2=\Theta \\
& \begin{array}{c}
W=M B(1-\cos \Theta) \\
\text { if } \Theta_1=90^{\circ} \text { and } \Theta_2=\Theta \\
W=-M B \cos \Theta
\end{array}
\end{aligned}
$

Potential Energy of a dipole kept in a magnetic field-
As $\Delta U=-W_{m a g}=W$
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 $\Delta U=M B\left(\cos \Theta_1-\cos \Theta_2\right)$

$
\begin{aligned}
& \text { if } \Theta_1=90^{\circ} \text { and } \Theta_2=\Theta \\
& \Delta U=U_{\theta_2}-U_{\theta_1}=U_\theta-U_{90}=-M B \cos \Theta
\end{aligned}
$

Assuming $\Theta_1=90^{\circ}$ and $U_{90^{\circ}}=0$
we can write $U=U_\theta=-\vec{M} \cdot \vec{B}$

Equilibrium of Dipole-

1. Stable Equilibrium-

$
\begin{aligned}
& \Theta=0^{\circ} \\
& \tau=0 \\
& U_{\min }=-M B
\end{aligned}
$

2. Unstable Equilibrium-

$
\begin{aligned}
& \Theta=180^{\circ} \\
& \tau=0 \\
& U_{\max }=M B
\end{aligned}
$

3. Not in equilibrium-

$
\begin{aligned}
& \Theta=90^{\circ} \\
& \tau_{\max }=M B \\
& U=0
\end{aligned}
$