Question

A 23.6 cm -diameter coil consists of 21 turns of circular copper wire 3.0 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.60×10?3 T/s.

Part A: Determine the current in the loop.

Part B: Determine the rate at which thermal energy is produced.

Answer 1

In what follows, D denotes the diameter of the coil, d denotes the diameter of the wire that made up the coil.

The resistivity of copper at 25°C is ? = 1.7×10^-8 ?.m
The length of the copper wire is
L = N (?D) = 21?×23.6cm = 495.6 ? cm = 4.956 ? meters
The sectional area of the wire is
A = ? d²/4 = (3.0²/4)? mm² = 2.25 ? mm² = 2.25×10^(-6) ? m²
The resistance of the coil
R = ?L/A = 1.7×10^-8×4.956 ? / 2.25×10^(-6) ? = 3.744×10^-2 ohm

The inductive electromotive force in the coil:
? = N d?/dt = NS dB/dt = (N?D²/4)dB/dt = 21?×(23.6×10?²)²×8.60×10^-3/4
? = 7.9000×10^-3 volt

(a) The current in the loop:
I = ? / R = 7.900×10^-3 / 3.744×10^-2 = 0.211 A

(b) The rate at which thermal energy is produced:
P = RI² = ?I = 7.900×10^-3× 0.211 = 1.666*10^-3 W