A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is n and the cross-sectional area of the coil is A. When the coil turns through 180° about its diameter, the charge flowing through the coil is Q. The total resistance of the circuit is R. What is the magnitude of the magnetic induction?

Question

A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is n and the cross-sectional area of the coil is A. When the coil turns through 180° about its diameter, the charge flowing through the coil is Q. The total resistance of the circuit is R. What is the magnitude of the magnetic induction? (A) (QR/nA) (B) (2QR/nA) (C) (QR/2RA) (D) (QR/2nA)

Tardigrade Admin · Accepted Answer

Induced charge Q=-(n B A/R)( cos θ2- cos θ1) =-(n B A/R)( cos 180°- cos 0°) ⇒ B=(Q R/2 n A)

$\frac{QR}{n A}$

$\frac{2 QR}{n A}$

$\frac{QR}{2 R A}$

$\frac{QR}{2 n A}$

Induced charge $Q = - \frac{n B A}{R} (cos θ_{2} - cos θ_{1})$
$= - \frac{n B A}{R} (cos 18 0^{\circ} - cos 0^{\circ})$
$\Rightarrow B = \frac{QR}{2 n A}$

nAQR​

nA2QR​

2RAQR​

2nAQR​

Induced charge Q=−RnBA​(cosθ2​−cosθ1​) =−RnBA​(cos180∘−cos0∘) ⇒B=2nAQR​

$\frac{QR}{n A}$

$\frac{2 QR}{n A}$

$\frac{QR}{2 R A}$

$\frac{QR}{2 n A}$

Induced charge $Q = - \frac{n B A}{R} (cos θ_{2} - cos θ_{1})$
$= - \frac{n B A}{R} (cos 18 0^{\circ} - cos 0^{\circ})$
$\Rightarrow B = \frac{QR}{2 n A}$