Q. The magnetic needle of a tangent galvanometer is deflected at angle of $30{^\circ}$ due to a current in its coil. The horizontal component of earth’s magnetic field is $0.34 \times 10^{- 4}\ T,$ then magnetic field at the center of the coil due to current
A) $1.96 \times 10^{- 5}\ T$
B) $1.96 \times 10^{- 4}\ T$
C) $1.96 \times 10^{4}\ T$
D) $1.96 \times 10^{5}\ T$
Explanation
(a) Given that, the horizontal component of earth’s magnetic field $B_{H} = 0.34 \times 10^{- 4}\ T$
$\theta = 30{^\circ}$
We know that, for tangent galvanometer
$\ \ \ \ \ \ B = B_{H}\tan\theta$
$\Rightarrow \ B = 0.34 \times 10^{- 4} \times \tan{30{^\circ}}$
$\ \ \ \ \ \ \ \ \ \ = 1.96 \times 10^{- 5}\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.