Q. A coil of $40\ \Omega$ resistance has 100 turns and radius $6\ mm$ is connected to ammeter of resistance of $160\ ohms$. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, $32\ \mu$ C charge flows through it. The intensity of magnetic field will be
A) $6.55\ T$
B) $5.66\ T$
C) $0.655\ T$
D) $0.566\ T$
Explanation
(d) $q = - \frac{N}{R}\left( B_{2} - B_{1} \right)A\cos\theta$
$32 \times 10^{- 6} = - \frac{100}{(160 + 40)}(0 - B) \times \pi \times \left( 6 \times 10^{- 3} \right)^{2} \times \cos{0{^\circ}}$
$\Rightarrow B = 0.565\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.