Question

A long straight wire of radius $a$ carries a steady current $i$. The current is uniformly distributed in the cross section of the wire. The ratio of the magnetic field at $\frac{a}{2}$ and $2a$ is:

• A. $\frac{1}{4}$
• B. $4$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $1$

Magnetic field due to a current carrying wire of radius a is given as,

$B=\frac{{\mu }_{0}Ix}{2\pi a^2}x<a$

$B=\frac{{\mu }_{0}I}{2\pi x}x\ge a$

$\therefore \frac{B_{x_1}}{B_{x_2}}=\frac{B_{a/2}}{B_{2a}}=\frac{x_1}{\left(1/x_a\right)}$ $(\because x_1=a/2,x_2=2a)$

$\frac{B_{a/2}}{B_{2a}}=x_1{x}_2$

$=\left(\frac{a}{2}\right)\times 2a$

$=1$

$\therefore \frac{B_{a/2}}{B_{2a}}=1$