Question

A long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distribute over its cross-section. The ratio of the magnetic fields $B$ and $B^{\prime }$, at radial distances $\left(a/2\right)$ and $2a$ respectively, from the axis of the wire is:

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

Answer 1

Answer: (A)

Let $J$ be the surface current density . So, $I=J.\pi a^2$

Using Ampere's law the field at radial distance $a/2$: $B.2\pi \left(a/2\right)={\mu }_0{I}_{en}$

Here $I_{en}=$current enclosed by amperical loop of radius a/2 $=J.\pi (a/2{)}^2=I/4$

So, $B=\frac{{\mu }_{0}I}{4\pi a}$

And at radial distance $2a$: $B^{\prime }.2\pi \left(2a\right)={\mu }_0{I}_{en}$

Here $I_{en}=I$ as it is outside the wire.

Thus, $B^{\prime }=\frac{{\mu }_{0}I}{4\pi a}$

$\therefore \frac{B}{B^{\prime }}=1$