A conducting circular coil with two turns sits on top of a uniform magnetic fiel

Question

A conducting circular coil with two turns sits on top of a uniform magnetic field with B = 8.0 mT, pointing out of the page, and the ends of it are attached to a resistor with R = 2.0 k ohm. The coil expands at a constant rate from a radius of 0.5 m to 1.0 m in 3 seconds. (a) While it is doing this, what is the direction of the current around the coil, and its value? How much total energy is dissipated through the resistor? (c) Suppose that instead the coil does not expand, but we change the magnetic field. How do we change it to get the same current?

Explanation / Answer

a) conter clokwise direction

b) emf = BdA/dt = 8*10^-3*3.14*(1^2-0.5^2)/3 = 0.00628 V

i = emf/R = 0.00628/2000 = 0.00314 mA

energy = p*t = i^2R*t = 0.059 J

c) now emf = AdB/dt

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