Q. 34.1( 40 Votes )

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Answer :


No. of turns in the solenoid coil, N = 15 turns/cm = 15000 turns/m

( 1m = 100 cm)

no. of turns in the solenoid coil per unit length, n = 15000 turns

Area of the solenoid coil = 2.0 cm2

In m2, Area will be 2 × 10 - 4 m2

Since current carried by the solenoid coil changes from 2.0 to 4.0 A in 0.1 seconds

Therefore, change in the current in the coil of solenoid = final current – initial current

di = 4.0 – 2.0 = 2.0 A

The change in time “dt” is given as ,dt = 0.1 seconds

Applying Faraday’s law, induced e.m.f can be calculated as follows:


Where Ф is flux induced in the loop

And, flux is given as, Ф = BA

Where “B” is the magnetic field and “A” is the area of the loop.

For a solenoid, B= μ0nI

Therefore, the equation (1) becomes:


Or E = Aμ0 n (dI/dt)…(2)

A, μo and n are constants

Substituting the values in equation (2), we get,

E = 7.54 × 10 - 6 V

Or E = 7.54 μV

( 10-6 = μ)

Hence, the induced voltage in the loop is 7.54 μV

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