An aluminum ring of radius 5.0 cm and resistance 4.0 10-4 is placed around the t

Question

An aluminum ring of radius 5.0 cm and resistance 4.0 10-4 is placed around the top of a long air-core solenoid with 1004 turns per meter and a smaller radius of 3.0 cm, as in Figure P20.62. If the current in the solenoid is increasing at a constant rate of 271 A/s, what is the induced current in the ring? Assume that 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 that the solenoid produces a negligible field outside its cross-sectional area.
(answer in Amperes) An aluminum ring of radius 5.0 cm and resistance 4.0 10-4 is placed around the top of a long air-core solenoid with 1004 turns per meter and a smaller radius of 3.0 cm, as in Figure P20.62. If the current in the solenoid is increasing at a constant rate of 271 A/s, what is the induced current in the ring? Assume that 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 that the solenoid produces a negligible field outside its cross-sectional area. (answer in Amperes)

Explanation / Answer

r = 5.0 cm, R = 4.0*10^-4 ohm, n = 1004 turns/m, r' = 3.0 cm, dI/dt = 271 A/s, mu_0 = 4*pi*10^-7 find the induced current i in the ring. the magnetic field produced by the solenoid over the area at the end of the solenoid B = 0.5 * mu_0*n*I magnetic flux in the ring = B*pi*r'^2 emf = d(B*pi*r'^2)/dt = d(0.5*mu_0*n*I*pi*r'^2)/dt = 0.5*mu_0*n*pi*r'^2 * dI/dt so i = emf/R = 0.5*mu_0*n*pi*r'^2/R * dI/dt = 1.21 A

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