A solenoid has an area A= 20 cm2, its length is 1= 2m and number of turns N= 300

Question

A solenoid has an area A= 20 cm2, its length is 1= 2m and number of turns N= 300 turns. What is the self-inductance of the solenoid? The current in solenoid is I=2A with direction as shown. What is the total energy of the magnetic field established inside the solenoid, within its volume? The solenoid is placed inside a larger, single-turn coil with area A'= 100cm2. The current in solenoid is turned off such that it goes from 2 A to 0A in 0.1 seconds. What is the current induced in the larger single coil during this 0.1 seconds interval. Show its direction.

Explanation / Answer

inductance of solenoid = L = uo*N^2*A/L

L = (4*pi*10^-7*300^2*20*10^-4)/2

L = 1.13*10^-4 H

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b)

energy stored = 0.5*L*I^2 = 0.5*1.13*10^-4*2^2 = 2.26*10^-4 J

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c)

flux through the coil = B*A = uo*N/L*i*A

emf induced = rate of change in flux = uo*(N/L)*A*(di/dt)

e = 4*pi*10^-7*(3000/2)*20*10^-4*2/0.1 = 7.54*10^-5 V

induced current = i = e/R = 2.51*10^-4 A <<---answer

direction counter clock wise

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