Question

An air-cored solenoid, 40 cm long and of cross-sectional area 5 cm², is tightly wound with 400 turns of copper wire and carries a steady current of 10 A. (a) Calculate the self-inductance of the solenoid. (b) Find the emf induced if the current in the solenoid decreases to zero in 0.2 s.

Answer 1

Data: l = 0.4 m, A = 5 × 10^-4 m², N = 400, I_i = 10 A, I_f = 0,

Δt = 0.2 s, µ₀ = 4π × 10⁻⁷ H/m

(a) Self-inductance of the solenoid,

L = `µ_0(N^2/l) A`

= `(4π × 10^-7) (400^2/0.4) (5 × 10 ^-4)`

= `(5π xx 10 ^-10) (16 xx 10^4) = 8π xx 10^-5`

= 8 × 3.142 × 10⁻⁵

= 2.514 × 10⁻⁴ H

(b) The rate of change of current,

`(dl)/dt = (I_f - I_i)/(Δt)`

= `(0 - 10)/0.2`

= − 50 A/s

∴ The emf induced is

e = `-L (dI)/(dt)`

= − (2.514 × 10⁻⁴) (−50)

= 125.7 × 10⁻⁴ V

= 12.57 mV