Question

A long straight wire of radius \(a\) carries a steady current \(I\). The current is uniformly distributed across its cross section. The ratio of the magnetic field at \(\frac{a}{2}\) and \(2a\) from axis of the wire is:

• A. 1 : 4
• B. 4 : 1
• C. 1 : 1
• D. 3 : 4

Answer 1

The Correct Option is C

For a point inside the wire (\(r<a\)), the magnetic field is given by:
\[B_1 \times 2\pi \frac{a}{2} = \mu_0 \frac{I}{4} \implies B_1 = \frac{\mu_0 I}{4\pi a}.\]
For a point outside the wire (\(r>a\)), the magnetic field is:
\[B_2 \times 2\pi \times 2a = \mu_0 I \implies B_2 = \frac{\mu_0 I}{4\pi a}.\]
The ratio of magnetic fields:
\[\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 : 1.\]