Chapter 30, Problem 031 If 52.7 cm of copper wire (diameter = 0.960 mm, resistiv

Question

Chapter 30, Problem 031 If 52.7 cm of copper wire (diameter = 0.960 mm, resistivity = 1.69 x 10-aQ-m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.81 mT/s, at what rate is thermal energy generated in the loop? Number the tolerance is +/-2% Click if you would like to Show Work for this question: Qen Show Work Units SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM Question Attempts: Unlimited SAVE FOR LATERSUBMIT ANSWER SUBMIT ANSWER

Explanation / Answer

Length of wire L = Circumference of loop

L = 2*pi*r

r = L/2*pi = 52.7/(2*pi)

r = 8.39 cm

A = pi*r^2 = pi*0.0839^2 = 2.21*10^-2 m^2

EMF is given by:

EMF = N*d(phi)/dt

EMF = N*A*dB/dt

EMF = 1*(2.21*10^-2 m^2)*8.81*10^-3 T/sec

EMF = 1.95*10^-4 V

Now Power is given by:

P = V^2/R

R = rho*L/A

R = 1.69*10^-8*0.527/(pi*(0.48*10^-3)^2) = 0.0123 ohm

So,

P = (1.95*10^-4)^2/0.0123 = 3.09*10^-6 W

Energy generated = 3.09*10^-6 J/sec

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