A magnetic dipole is placed under the effect of two magnetic fields inclined at $75^{0}$ to each other one of the fields has a magnetic induction of $1.5 X 10^{- 2} T$ . The magnet comes to rest at an angle of $30 °$ with the direction of this field, the magnitude of the other field is

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$1.5 X 10^{- 2} T$

$1.5 \sqrt{2} X 10^{- 2} T$

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

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

answer is C.

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Detailed Solution

At the equilibrium position

$M B_{1} \sin θ_{1} = M B_{2} \sin θ_{2}$

Given $θ_{1} = 30 ° a n d θ_{2} = 75 ° - 30 ° = 45 °$

$\begin{matrix} B_{1} = 1.5 \times 10^{- 2} T B_{2} = ? \\ ∴ B_{2} = \frac{B_{1} \sin θ_{1}}{\sin θ_{2}} \end{matrix} \Rightarrow B_{2} = \frac{1.5 \times 10^{- 2} \times \sin 30 °}{\sin 45 °} = \frac{1.5 \times 10^{- 2} \times 1 \times \sqrt{2}}{2} \Rightarrow B_{2} = \frac{1.5 \times 10^{- 2}}{\sqrt{2}} T$