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

Given,

The magnitude of the first magnetic field, B1 = 1.2 × 10–2 T

The angle between the magnetic field directions, $\dpi{100} \theta$ = 60°

The angle between the dipole and the magnetic field $\dpi{100} B_{1}$ is $\dpi{100} \theta_{1}$ = 15°

Let Bbe the magnitude of the second magnetic field and M be the magnetic dipole moment

Therefore, the angle between the dipole and the magnetic field B2 is $\dpi{100} \theta_{2}$ = $\dpi{100} \theta - \theta_{1}$= 45°

Now, at rotational equilibrium,

The torque due to field B1 = Torque due to field B2

$MB_{1}sin\theta_{1}= MB_{2}sin\theta_{2}$

$B_{2} = \frac{MB_{1}sin\theta_{1}}{Msin\theta_{2}} = \frac{1.2\times10^{-2}\times sin15\degree}{sin45\degree}$

$= 4.39 \times$ $\dpi{100} 10^{-3}$$T$

Hence the magnitude of the second magnetic field $= 4.39 \times$$\dpi{100} 10^{-3}$$T$

Exams
Articles
Questions