Question

A long straight wire of radius x carries a steady current I. The current is uniformly distributed across its cross section. Calculate the ratio of the magnetic field at \(\frac {x}{4}\) and 8x.

(a) 4

(b) 3

(c) 2

(d) 1

I had been asked this question in unit test.

I want to ask this question from Ampere’s Circuital Law in chapter Moving Charges and Magnetism of Physics – Class 12

Answer 1

The correct answer is (c) 2

Easiest explanation: The magnetic field due to a long straight wire of radius x carrying a current I at a point distant r from the axis of the wire is given by:

Bin = \(\frac {\mu_o Ir}{2\pi x^2}\)   (r < x)

Bin = \(\frac {\mu_o I}{2\pi r}\)   (r > x)

The magnetic field at a distance of r = \(\frac {x}{4}\):

B1 = \(\frac {\mu_o I (\frac {x}{4})}{2\pi x^2} = \frac {\mu_o I}{8\pi x}\)

The magnetic field at a distance of r = 8x:

B2 = \(\frac {\mu_o I}{2\pi (8x)} = \frac {\mu_o I}{16\pi x}\)

Therefore, \(\frac {B1}{B2} = \frac {\frac {\mu_o I}{8\pi x}}{\frac {\mu_o I}{16\pi x}}\)

\(\frac {B1}{B2} = \frac {16}{8}\) = 2