Question

A circular coil of 200 turns has a radius of 10cm and carries a current of 2.0A. (a) Find the magnitude of the magnetic field vector B at the centre of the coil, (b) At what distance from the centre along the axis of the coil will the field B drop to half its value at the centre? (3√4 = 1.5874 ...)

Answer 1

n = 200, i = 2A, r = 10cm = 10 × 10–2n

= 2 × 4 × 3.14 × 10–4 = 25.12 × 10–4T = 2.512mT

=>(a2 +d2)3/2 2a3 => a2 + d2 = (2a3)2/3

=> a2 + d2 = (21/3 a)2 => a2 + d2 =22/3a2 => (10–1)2 + d2 = 22/3 (10–1)2

=> 10–2 + d2 = 22/3 10–2 => (10–2)(22/3 – 1) = d2 => (10–2) (41/3 – 1) = d2

=> 10–2(1.5874 – 1) = d2 => d2 = 10–2 × 0.5874

= 10–1 × 0.766m = 7.66 × 10–2 = 7.66cm.