A circular coil of 500 turns of wire has an enclosed area of 0.1 m 2 per turn. It is kept perpendicular to a magnetic field of induction 0.2 T and rotated by 180° about a diameter perpendicular to the field in 0.1 s. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of 50 ohms?

Question

A circular coil of 500 turns of wire has an enclosed area of 0.1 m 2 per turn. It is kept perpendicular to a magnetic field of induction 0.2 T and rotated by 180° about a diameter perpendicular to the field in 0.1 s. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of 50 ohms? (A) 0.2 C (B) 0.4 C (C) 2 C (D) 4 C

Tardigrade Admin · Accepted Answer

Δ Q =(N B A/R)( cos θ1- cos θ2) =(500 × 0.2 × 0.1( cos 0°- cos 180°)/50)=0.4 C

0.2 C

0.4 C

2 C

4 C

ΔQ=RNBA​(cosθ1​−cosθ2​) =50500×0.2×0.1(cos0∘−cos180∘)​=0.4C

$Δ Q = \frac{NB A}{R} (cos θ_{1} - cos θ_{2})$
$= \frac{500 \times 0.2 \times 0.1 ( c o s 0 ^{\circ} - c o s 18 0 ^{\circ} )}{50} = 0.4 C$