Question

A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle, of ${75}^{\circ }$. One of the fields has a magnitude of $\sqrt{2}\times {10}^{-2}T$. The dipole attains stable equilibrium at an angle of ${30}^{\circ }$ with this field. What is the magnitude of the other field?

• A. 0.01 T
• B. 0.02 T
• C. 0.03 T
• D. 0.04 T

Answer 1

The correct option is A

0.01 T

Refer to Fig. Let ${\theta }_1(={30}^{\circ })$ be the angle between the magnetic moment vector m and the field vector $B_1(=1.5\times {10}^{-2}T)$. Then, as shown in Fig. the angle between m and the other field $B_2$ will be ${\theta }_2={75}^{\circ }-{30}^{\circ }={45}^{\circ }$
The field $B_1$ exerts a torque ${\tau }_1=m\times B_1$ on the dipole and the field $B_2$ exerts a torque ${\tau }_2=m\times B_2$, where m in the magnetic moment of the dipole. Since the dipole is in stable equilibrium, the net torque $\tau (={\tau }_1+{\tau }_2)$ must be zero, i.e. the two torques must be equal and opposite. In terms of magnitudes, we have $m{B}_{1}sin{\theta }_1=m{B}_{2}sin{\theta }_2$
or $B_2=\frac{B_{1}sin{\theta }_1}{sin{\theta }_2}=\frac{\sqrt{2}\times {10}^{-2}\times sin{30}^{\circ }}{sin{45}^{\circ }}=0.01T$