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 2A 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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