Q. The radius of the coil of a tangent galvanometer which has 10 turns is 0.1 m. The current required to produce a deflection of $60{^\circ}\ \left( B_{H} = 4 \times 10^{- 5}\ T \right)$ is
A) 3 A
B) 1.1 A
C) 2.1 A
D) 1.5 A
Explanation
(b) Current, $I = B_{H} \times \frac{2r{\ tan}\theta}{n\mu_{0}}$
$= 4 \times 10^{- 5} \times \frac{2 \times 0.1 \times \tan{60{^\circ}}}{10 \times 4\pi \times 10^{- 7}} = 1.1\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.