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

Answer: (C)

1 : 1

Answer 2

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:
$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 : 1.$