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

N = 100,

R = 8 × 10⁻² m,

I = 0.4 A,

μ₀ = 4π × 10⁻⁷ T m/A

B = `(mu_0 "NI")/"2R"`

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

= 3.14 × 10⁻⁴ T

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