Question

The distance at which the magnetic field on axis as compared to the mag. field at the centre of the coil carrying current I and radius R is (1/8), would be

(a) R

(b) √2 R

(c) 2R

(d) √3 R

Answer 1

The correct option (d) √3 R

Explanation:

mag. field at distance x from the centre of coil having radius R is

B = [(μ0R2I)/{2(R2 + x2)3/2}]

at centre x = 0 hence B = (μ0I/2R)

Let distance required is r hence at x = r,

B' = [(μ0R2I)/{2(R2 + r2)}3/2]

given B' = (1/8)B

∴ [{(μ0R2I)/[2(R2 + r2)3/2]}/{(μ0I)/2R}] = (1/8)

∴ [(R2 × R)/{(R2 + r2)3/2}] = 1/8

8R3 = (R2 + r2)3/2

∴ 8 = [(R2 + r2)3/2/(R2)3/2] = [(R2 + r2)/R2]3/2

∴ Squaring both sides,

64 = [(R2 + r2) / (R2)]3 i.e. 43 = [(R2 + r2)/(R2)]3

∴ Taking cuberoots

4 = [(R2 + r2)/R2]

∴ 3R2 = r2

r = √3 R