Question

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 1

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m

Number of turns per unit length, n = 1500 turns

The solenoid has a small loop of area, A = 2.0 cm²

= 2 × 10⁻⁴ m²

Current carried by the solenoid changes from 2 A to 4 A.

∴ Change in current in the solenoid, di = 4 − 2 = 2 A

Change in time, dt = 0.1 s

Induced emf in the solenoid is given by Faraday’s law as:

`"e" = ("d"phi)/("dt")` ........(1)

Where,

`phi` = Induced flux through the small loop = BA ...........(2)

B = Magnetic field = `mu_0"ni"` .....(3)

μ₀ = Permeability of free space

= 4π × 10⁻⁷ H/m

Hence, equation (1) reduces to:

`"e" = "d"/("dt")("BA")`

= `"A"mu_0"n" xx (("di")/("dt"))`

= `2 xx 10^-4 xx 4pi xx 10^-7 xx 1500 xx 2/0.1`

= 7.54 × 10⁻⁶ V

Hence, the induced voltage in the loop is 7.54 × 10⁻⁶ V.