A 238 cm diameter coil consists of 20 turns of circular copp

A 23.8 cm -diameter coil consists of 20 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 8.12×10³ T/s .

Determine the current in the loop and the rate at which thermal energy is produced.

Solution

here,

diameter of coil , D = 0.238 m

radius of coil , R = 0.119 m

radius of wire , r = 0.0015 m

channge of rate of magnetic feild , dB/dt = 8.12 * 10^-3 T/s

resistance , R = p * L /area

R = p *( 2 * pi * R * n) /( pi * r^2)

R = 1.68 * 10^-8 * 2 * 0.119^2 * 20 / 0.0015^2

R = 4.23 * 10^-3 ohm

EMF induced, V = area * dB/dt

V= pi * 0.119^2 * 8.12 * 10^-3

V = 0.361 * 10^-3 V

the current in the loop , I = V/R

I = 0.085 A

the current in the loop is 0.085 A

the rate at which thermal energy produced, P = I^2 * R

P = 0.085^2 * 4.23 * 10^-3

P = 3.08 * 10^-5 W