Question

A long, straight wire, of radius $a$, carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire, at distances $\frac{a}{3}\text{and}2a,$ respectively from the axis of the wire, is:

• A. $\frac{2}{3}$
• B. $2$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $\frac{2}{3}$

Let $a$ be the radius of the wire.

Magnetic field at point $A$ (inside the wire):
$B_A=\frac{{\mu }_{0}ir}{2\pi a^2}=\frac{{\mu }_{0}i\left(\frac{a}{3}\right)}{2\pi a^2}=\frac{{\mu }_{0}i}{6\pi a}$


Magnetic field at point B (outside the wire):

$B_B=\frac{{\mu }_{0}i}{2\pi \left(2a\right)}=\frac{{\mu }_{0}i}{4\pi a}$

$\therefore \frac{B_A}{B_B}=\frac{\left(\frac{{\mu }_{0}i}{6\pi a}\right)}{\left(\frac{{\mu }_{0}i}{4\pi a}\right)}=\frac{4}{6}=\frac{2}{3}$

Hence, $\left(A\right)$ is the correct answer.