Q. A tangent galvanometer has a coil of 25 turns and a radius of 15 cm. The horizontal component of the earth’s magnetic field is $3 \times 10^{- 5}$T. The current required to produce a deflection of 45${^\circ}$ in it is
A) 0.29 A
B) 0.14 A
C) 1.2 A
D) $3.6 \times 10^{- 5}$ A
Explanation
(a) $n = 25,r = 15\ cm = 15 \times 10^{- 2}\ m\mathbf{,}$
$H = 3 \times 10^{- 5}\ T,\ i = ?\ \theta = 45{^\circ}$
From $F = \frac{\mu_{0}n_{i}}{2r} = H\tan\theta$
$i = \frac{2rH\ tan\theta}{\mu_{0}n} = 2 \times \frac{15 \times 10^{- 2} \times 3 \times 10^{- 5}\tan 45{^\circ}}{(4\pi \times 10^{- 7}) \times 25}$
$\ = 0.29$ A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.