Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius `'R'` and outer radius `'2R'`, cylinder carrying uniformly distributed current along it's axis. The magnetic induction at point `'P'` at a distance `(3R)/2` from the axis of the cylinder will be

Options

A. Zero
B. `(5 mu_(0) i)/(72 pi R)`
C. `(7mu_(0) i)/(18 pi R)`
D. `(5 mu_(0) i)/(36 pi R)`

Answer

A

Explanation

By using `B = (mu_(0) i)/(2pi r) ((r^(2) - a^(2))/(b^(2) - a^(2)))` , here `r = (3R)/(2)` a = R , b = 2 R
We get `B = (mu_(0))/(2pi ((3R)/(2))) xx {((3R)/(2))^(2) - R^(2))/((2R)^(2) - R^(2))} = (5 mu_(0) i)/(36 pi R)`
Aliter : For `R lt x lt 2 R `
Current `I_("net") = ((i)/(pi (2R)^(2) - R^(2))) [ pi ((3R)/(2))^(2) - (R)^(2) pi ]`
= `(i)/(3R^(2)) [ (9R^(2))/(4) - R^(2)]`
`= (i)/(3R^(2)) xx (5R^(2))/(4) = (5i)/(12)`

Apply ampere circuit law - `oint B.dl = mu_(0) I_("net") implies B xx 2pi (3R)/(2) = (mu_(0) 5i)/(12) = (5mu_(0) i)/(36 pi R)`