Question

An iron core solenoid with 400 turns has a cross-section area of 4.0 cm². A current of 2.0 A passing through it produces B = 0.5 T. How large an induced emf is induced in it? If the current is turned off in 0.1 see, what is the self-inductance of the solenoid?

Answer 1

The induced emf in the solenoid is 0.8 V and the self-inductance of the solenoid is 16 H.

Step-by-step explanation:

The emf induced in the solenoid can be calculated using Faraday's Law of Electromagnetic Induction, which states that the induced emf is equal to the rate of change of magnetic flux. In this case, the magnetic flux is changing because the current is being turned off.

Induced emf, E = -dΦ/dt, where Φ = B.A.N. In this question, the magnetic field B = 0.5 T, the cross-sectional area A = 4.0 cm² (or 4.0 x 10^-4 m² in SI units), and N = 400 turns. Hence, Φ = B.A.N = 0.5 T x 4.0 x 10^-4 m² x 400 turns = 0.08 Wb (weber). If the current is turned off in time dt = 0.1 s, then the rate of change of Φ is -0.08 Wb / 0.1 s = -0.8 Wb/s. The negative sign indicates a decrease in magnetic flux. Hence, the induced emf is E = -(-0.8 V) = 0.8 V.

The self-inductance, L of the solenoid can be obtained using the formula L = N.Φ/I, where I = 2.0 A is the current. Hence, L = 400 turns x 0.08 Wb / 2.0 A = 16 H.

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