Question

A circular coil of radius 2.00 cm has 50 turns. A uniform magnetic field B = 0.200 T exists in the space in a direction parallel to the axis of the loop. The coil is now rotated about a diameter through an angle of 60.0°. The operation takes 0.100 s. (a) Find the average emf induced in the coil. (b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω, calculate the net charge crossing a cross-section of the wire of the coil.

Answer 1

Charge is product of Current and Time

Explanation:

Given:

Number of turns of the coil, N = 50

Magnetic field through the circular coil, $\vec{B}$} = 0.200 T

Radius of the circular coil, r = 2.00 cm = 0.02 m

Angle through which the coil is rotated, $\Theta$ = 60°

Time taken to rotate the coil, t = 0.100 s

• (a) The emf induced in the coil is given by

e = $- \frac {N \Delta \phi}{\Delta t} = \frac {N(\vec{B_f}\vec{A_f} - \vec{B_t}\vec{A_t})}{T}$

= $\frac {NBA(cos0 - cos60)}{T}$

= $\frac {50\times 2\times 10^-^1 \times \pi (0.02)^2}{2 \times 0.1}$

= $5 \times 4 \times 10^-^5 \times\pi$

= $2\pi \times10^-^2$ V = $6.28 \times 10^-^3$ V

• (b) The current in the coil is given by

i = $\frac {e}{R} = \frac {6.28 \times 10^-^3}{4}$

= $1.57 \times 10^-^3$ A

The net charge passing through the cross section of the wire is given by

Q = it = $1.57 \times 10^-^3 \times 10^-^1$

= $1.57 \times 10^-^4$ C