A 2.5 m-long solenoid is 14.0 cm in diameter and consists of 2500 turns of wire.
ID: 1693731 • Letter: A
Question
A 2.5 m-long solenoid is 14.0 cm in diameter and consists of 2500 turns of wire. The current in the solenoid is increasing at the rate of 1.1 kA/s.Find the current in a wire loop with diameter 7.00 cm and resistance 5.0 O, lying inside the solenoid in a plane perpendicular to the solenoid axis. Repeat for a similarly oriented loop with diameter 21.0 cm, lying entirely outside the solenoid. Find the current in a wire loop with diameter 7.00 cm and resistance 5.0 O, lying inside the solenoid in a plane perpendicular to the solenoid axis. Repeat for a similarly oriented loop with diameter 21.0 cm, lying entirely outside the solenoid.
Explanation / Answer
Given length of the solenoid L = 2.5 m diameter d = 14 cm, ==> radius r = 7 cm = 0.07 m The current in the solenoid is increasing at the rate of 1.1 kA/s.dI / dt = 1.1 k A / s magnetic field inside the solenoid B = µ_o N I / L induced emf in the wire e = d/dt ( B A ) = ( pr_1 ^2 ) ( dB / dt ) = ( pr_1 ^2 ) ( d(µ_o N I / L) / dt ) = (µ_o N / L) ( pr_1 ^2 ) ( dI /dt ) = 4p *10^-7 ) ( 2500 )(3.14*0.035^2 ) ( 1.1*10 ^3 ) / 2.5 = 5.3 mV current in the wire loop I = e / R = 5.3mV / 5 = 1.06 mA b) e = d/dt ( B A ) = ( pr_1 ^2 ) ( dB / dt ) = ( pr_1 ^2 ) ( d(µ_o N I / L) / dt ) = (µ_o N / L) ( pr_1 ^2 ) ( dI /dt ) = 4p *10^-7 ) ( 2500 )(3.14*0.105^2 ) ( 1.1*10 ^3 ) / 2.5 = 0.0478mV current in the wire loop I = e / R = 0.0478V/ 5 = 0.00956 A = ( pr_1 ^2 ) ( dB / dt ) = ( pr_1 ^2 ) ( d(µ_o N I / L) / dt ) = (µ_o N / L) ( pr_1 ^2 ) ( dI /dt ) = 4p *10^-7 ) ( 2500 )(3.14*0.105^2 ) ( 1.1*10 ^3 ) / 2.5 = 0.0478mV current in the wire loop I = e / R = 0.0478V/ 5 = 0.00956 A
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