Question

A long, straight metal has a long hole of radius a drilled parallel to the rod axis with cross-sectional view as shown. If the rod carries a current $I$, the value of the magnetic field on the axis of the rod is:

• A. $B=\frac{{\mu }_0{Ia}^2}{2\pi c(b^2-a^2)}$
• B. $B=\frac{{\mu }_{0}I}{2\pi c}$
• C. $B=\frac{{\mu }_0{Ia}^2}{4\pi c(b^2-a^2)}$
• D. $B=zero$

Answer 1

The correct option is A $B=\frac{{\mu }_0{Ia}^2}{2\pi c(b^2-a^2)}$

Suppose that cavity contains same the current and same -ve current which is overall zero.

as $J=\frac{I}{\pi (b^2-a^2)}$,

current in cavity $I_1=\frac{I\pi a^2}{\pi (b^2-a^2)}$

$I_1=\frac{I{a}^2}{(b^2-a^2)}$

Then now we will calculate the magnetic field due to a cylinder of 'b' radius with current (+ve) and another cylinder of radius a with current (-ve).(Here positive and negative only denote the direction of flowing current w.r.t each other.)

${\overrightarrow{B}}_0={\overrightarrow{B}}_{cylinder}+{\overrightarrow{B}}_{Cavity}$

$=\frac{{\mu }_0{I}_1}{2\pi d_1}+\frac{{\mu }_0{I}_2}{2\pi d_2}$

$d_1=0,d_2=c$

On keeping value of $d_1$and $d_2$

$B_O=0+\frac{\mu I{a}^2}{2\pi (b^2-a^2)}\frac{1}{C}=\frac{\mu I{a}^2}{2\pi C(b^2-a^2)}$

$B=\frac{{\mu }_{0}I{a}^2}{2\pi C(b^2-a^2)}$

Option A is correct.