Question

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is ${60}^{\circ }$, and one of the fields has a magnitude of $1.2\times {10}^{-2}T$ T. If the dipole comes to stable equilibrium at an angle of ${15}^{\circ }$ with this field, what is the magnitude of the other field?

Answer 1

Step 1: Find the angle between dipole and unknown magnetic field.

Given, Magnitude of one of the magnetic fields, $B_1=1.2\times {10}^{-2}T$

Angle between two fields, $\theta ={60}^{\circ }$

At stable equilibrium, the angle between the dipole and the field $B_1,{\theta }_1={15}^{\circ }$

Let the magnitude of the other magnetic field be $B_2$.

Angle between the dipole and the field $B_2$

${\theta }_2=\theta -{\theta }_1={60}^{\circ }-{15}^{\circ }={45}^{\circ }$

Step 2: Calculate magnitude of unknown magnetic field.

At rotational equilibrium, the torque due to both fields must balance each other

Torque due to field $B_1$= torque due to field $B_2$

$M{B}_1\sin {\theta }_1=M{B}_2\sin {\theta }_2$

$B_2=\frac{B_1\times \sin {\theta }_1}{\sin {\theta }_2}$

$B_2=\frac{1.2\times {10}^{-2}\times \sin {15}^{\circ }}{\sin {45}^{\circ }}$

$B_2=4.39\times {10}^{-3}T$

Final answer: $4.39\times {10}^{-3}T$