Question

A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux linked with each turn when it carries a current of 1 A.

Answer 1

Length of the solenoid, l = 50 cm = 50 × 10^-2 m

No. of turns/cm = 400

For 50 cm, No. of turns N = 400 × 50 = 20,000

Diameter of the solenoid = 0.04 m

∴ Radius of the solenoid = 0.02 m

Current passing through the solenoid = 1 A

Area of the solenoid = πr²

= 3.14 × 0.02 × 0.02 m²

Formula :-

Magnetic flux, φ = µ₀ n² AIl

n = `"N"/l`

`therefore φ = (mu_0 "N"^2 "AI")/l`

`φ = (4 xx 3.14 xx 10^-7 xx 20,000 xx 20,000 xx 3.14 xx 0.02 x 0.02 xx 1)/(50 xx 10^-2)`

`= (6310144 xx 10^-7)/(50 xx 10^-2)`

`= 126202.88 xx 10^-7 xx 10^2`

= 126202.88 × 10^-5

φ = 1.262 Wb