Question

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer 1

Given: N = 100

R = 8 × 10⁻² m

I = 0.4 A

μ₀ = 4π × 10⁻⁷ T m/A

Formula: B = `(mu_0 NI)/2R`

= `((4pi xx 10^-7)(100)(0.4))/(2(8 xx 10^-2))`

= `((4 xx 3.14 xx 10^-7)(100)(0.4))/(16 xx 10^-2)`

= `(5.024 xx 10^-7)/(16 xx 10^-2)`

= 0.314 × 10⁻⁵ T

= 3.14 × 10⁻⁴ T

Hence, the magnitude of the magnetic field is 3.14 × 10⁻⁴ T.