Question

# A 20.8 cm -diameter coil consists of 22 turns of circular copper wire 2.8 mm in...

A 20.8 cm -diameter coil consists of 22 turns of circular copper wire 2.8 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.50×10^?3 T/s .

a) Determine the current in the loop

b) Determine the rate at which thermal energy is produced.

Resistance of copper wire will be

R = rho*L/A

rho = 1.7*10^-8 ohm-m

L = 22*2*pi*10.4*10^-2 = 14.37 m

A = pi*d^2/4 = pi*(2.8*10^-3)^2/4 = 6.16*10^-6 m^2

R = 1.7*10^-8*14.37/(6.16*10^-6) = 0.0396 ohm

Emf is given by:

EMF = N*d(phi)/dt

EMF = N*A*dB/dt

Using given values:

EMF = 22*(pi*0.104^2 m^2)*(8.5*10^-3 T/sec)

EMF = 0.00635 V

Now Current will be

i = V/R = 0.00635/0.0396

i = 0.160 Amp

Part B

rate at which thermal energy produced will be

P = V*i = 0.00635*0.160

P = 0.001016 W = 1.016 mW

Please Upvote. Let me know if you have any doubt.

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