Q. A coil of 50 turns and area $1.25 \times 10^{- 3}\ m^{2}$ is pivoted about a vertical diameter in a uniform horizontal magnetic field and carries a current of 2 A. When the coil is held with its plane is $N - S$ of 2A. When the coil is held with its plane in $N - S$ direction, it experience a couple of 0.04 N-m; and when its plane is $E - W,$ the corresponding couple is 0.03 N-m. The magnetic induction is
A) 0.2 T
B) 0.3 T
C) 0.4 T
D) 0.5 T
Explanation
(c) $M = ni\ A = 50 \times 2 \times 1.25 \times 10^{- 3} = 0.125\ Am^{2}$
If normal to the face of the coil makes an angle $\theta$ with the magnetic induction $B$, then in 1st case, torque $= MB\ cos\theta = 0.04$, and in second case,
Torque $= MBsin\theta = 0.03$
$\therefore MB = \sqrt{(0.04)^{2} + (0.03)^{2}} = 0.05$
$B = \frac{0.05}{M} = \frac{0.05}{0.125} = 0.4$T
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.