Q. 34.1( 40 Votes )

# A long solenoid with 15 turns per cm has a small loop of area 2.0 cm^{2} 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 :

Given:

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 cm^{2}

In m^{2}, Area will be 2 × 10 ^{- 4} m^{2}

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:

…(1)

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= μ_{0}nI

Therefore, the equation (1) becomes:

Or

Or E = Aμ_{0} n (dI/dt)…(2)

∵ A, μ_{o} 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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A rod AB moves with a uniform velocity v in a uniform magnetic field, as shown in the figure.

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