Explanation

(d) Given $A = 2 \times 10^{- 2}\ m^{2}$
$N = 100$
$\ I = 5\ A$
When the coil is held with its plane in North-South direction then its torque
$\tau_{1} = 0.3\ Nm$
$\tau_{2} = MB\sin\theta$
When the plane is in East-West direction then it’s torque
$\tau_{2} = 0.4\ Nm$
$\tau_{2} = MB\sin(90{^\circ} - \theta)$
$\tau_{2} = MB\cos\phi$
$\therefore\frac{MB\sin\theta}{MB\cos\theta} = \frac{\tau_{1}}{\tau_{2}}$
$\tan\theta = \frac{3}{4}$
or $\theta = \tan^{- 1}\left( \frac{3}{4} \right)$
or $\theta = 370{^\circ}$
Then $\sin\theta = 0.6$
$B = \frac{\tau_{1}}{M\sin\theta}$
$\Rightarrow \ B = \frac{0.3}{NIA\sin\theta}$
$\Rightarrow \ B = \frac{0.3}{100 \times 5 \times 2 \times 10^{- 2} \times 0.6}$
$B = 0.05\ T$