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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) 1I 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»

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


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