Question

A rectangular coil of length 0.12 m and with. 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/m${}^2$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30 with the direction of the field, the torque required to keep the coil in stable equilibrium will be

• A. 0.12 Nm
• B. 0.15 Nm
• C. 0.20 Nm
• D. 0.24 Nm

Answer 1

The correct option is C 0.20 Nm

	$\text{Given}$ : N = 50, I =2A , A = 0.12 $\times 0.1m=0.012m^2$ $B=0.2Wb/M^2$ $\text{To find}$ : torque $\text{Solution}$ : The required torque is $t=NIAB\sin \theta ;$ where N is the number of turns in the coil, I is the current through the coil, B is the uniform magnetic field, A is the area of the coil and & $\theta$ is the angle between the direction of the magnetic field and normal to the plane of the coil.
	Therefore. $\theta ={90}^{\circ }-{30}^{\circ }={60}^{\circ }$
	$\therefore t=\left(50\right)\left(2A\right)\left(0.012M^2\right)\left(0.2Wb/m^2\right)sin{60}^{\circ }$
	= 0.20 Nm $\text{Hence C is the correct option}$