An aluminum ring of radius 5.0 cm and resistance 5.0 * 10-4 Ohm is placed around

Question

An aluminum ring of radius 5.0 cm and resistance 5.0 * 10-4 Ohm is placed around the top of a long air-core solenoid with 997 turns per meter and a smaller radius of 3.0 cm, as in the figure below. If the current in the solenoid is increasing at a constant rate of 272 As., what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

Explanation / Answer

field at center of solenoid = u n I

field at end of solenoid is half this, or B = u n I / 2

mag flux in the aluminum ring = B * area of solenoid = (u n I /2) * pi * r^2 where r is radius of solenoid

induced current in ring = (dB/dt) / R where R is resistance

or

induced current = u n dI/dt * pi r^2 / 2R =

= 4pi x10^-7 * 997 * 272 * pi * 0.03^2 / 2 * 0.0005 =

= 0.9635 A

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