A long straight metal rod has a very long hole of radius `'a'` drilled parallel to the rod axis as shown in the figure. If the rod carries a current `I`, find the magnetic field on axis of hole. Given `C` is the centre of the hole and `OC=c`.

Options

A. `(mu_(0)ic)/(pi(b^(2)-a^(2)))`
B. `(mu_(0)ic)/(2pi(b^(2)-a^(2)))`
C. `(mu_(0)I(b^(2)-a^(2)))/(2pic)`
D. `(mu_(0)ic)/(2pib^(2))`

Answer

B

Explanation

In the rod current density
`j=i/(pi(b^(2)-a^(2)))`
Actual field is the vector sum of two current carrying rods, having same current density but in opposite direction. On the hole axis, only the larger rod contributes magnetic field. Imaging an amperian, loop of radius `E` and apply Ampere law.

`ointvec(B).dvec(l)=mu_(0)i_(enclosed)`
`B2pic=mu_(0)(jpic^(2))`
`B=(mu_(0)ipic^(2))/(2picx(b^(2)-a^(2)))=(mu_(0)ic)/(2pi(b^(2)-a^(2)))`
`B=(mu_(0)ic)/(2pi(b^(2)-a^(2)))`