Question

A coil of area $100{\text{cm}}^2$ having $500$ turns carries a current of $1\text{mA}$. It is suspended in a uniform magnetic field of ${10}^{-3}{\text{wb /m}}^2$. If the torque acting on the coil is $250\times {10}^{-8}\text{Nm}$ , then the angle made by the plane of the coil with the field is

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${60}^o$

$i=1\text{mA}={10}^{-3}\text{A};N=500;A=100\times {10}^{-4}{\text{m}}^2$

$B={10}^{-3}{\text{Wb/m}}^2;\tau =250\times {10}^{-8}\text{Nm}$

$\varphi$ is the angle made by the axis of the coil with the field.

$\tau =BiAN\sin \varphi$

$250\times {10}^{-8}={10}^{-3}\times {10}^{-3}\times 100\times {10}^{-4}\times 500\times \sin \varphi$

$\Rightarrow \sin \varphi =\frac{25}{50}$

$\Rightarrow \varphi ={30}^o$

But the plane of the coil makes an angle $\theta$ with the field and $\theta ={90}^{\circ }-\varphi$

$\therefore \theta =90-\varphi =90-30$

$\Rightarrow \theta ={60}^o$

Hence, option (c) is the correct answer.