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A uniform magnetic field acts perpendicular to the plane containing a circular copper ring of diameter D = 2 m. The wire forming the ring has a circular cross-section of diameter d = 2 mm. The resistivity of copper is rho = 10⁻⁸ ohm-m. At what rate must the magnetic field change with time so that the induced current in the ring is 10 A?

1. 4/(5*pi) T/s
2. 2/(5*pi) T/s
3. 1/(5*pi) T/s
4. none of these

Correct answer: 1/(5*pi) T/s

Solution

The ring's resistance is R = rho*(2*pi*R_ring)/(pi*(d/2)²) = 0.02 ohm. The induced EMF is (dB/dt)*pi*(1)². Setting EMF = I*R gives dB/dt = 10*0.02/pi = 1/(5*pi) T/s.

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