Question

A magnetic dipole is placed under the effect of two magnetic fields inclined at ${75}^{\circ }$ to each other. One of the fields has a magnetic induction of $1.5\times {10}^{-2}T$. The magnet comes to rest at an angle of ${30}^o$ with the direction of this field. The magnitude of the other field is:

• A. $1.5\times {10}^{-2}T$
• B. $1.5\sqrt{2}\times {10}^{-2}T$
• C. $\frac{1.5}{\sqrt{2}}\times {10}^{-2}T$
• D. $\frac{1.5}{2\sqrt{2}}\times {10}^{-2}T$

Answer 1

Answer: (C)

$\frac{1.5}{\sqrt{2}}\times {10}^{-2}T$

The components of both the field in the direction of dipole must be equal in magnitude. Since the other field is at an angle of ${75}^0-{30}^0={45}^0$ with the dipole. Hence,

$1.5\times {10}^{-2}sin\left({30}^0\right)=Bsin\left({45}^0\right)$

$B=\frac{1.5}{\sqrt{2}}\times {10}^{-2}T$