Q. A cylindrical conductor of radius ‘$R$’ carries a current ‘$i$’. The value of magnetic field at a point which is $R/4$ distance inside from the surface is $10T$. Find the value of magnetic field at point which is $4R$ distance outside from the surface
A) 4/3 $T$
B) 8/3 $T$
C) 40/3 $T$
D) 80/3 $T$
Explanation
(b) Magnetic field inside the cylindrical conductor
$B_{in} = \frac{\mu_{0}}{4\pi}.\frac{2ir}{R^{2}}\ (R =$ Radius of cylinder, $r = \ $distance of observation point from axis of cylinder)
Magnetic field out side the cylinder at a distance $r'$ from it’s axis $B_{out} = \frac{\mu_{0}}{4\pi}.\frac{2i}{r^{'}}$
$\Rightarrow \frac{B_{in}}{B_{out}} = \frac{{rr}^{'}}{R^{2}} \Rightarrow \frac{10}{B_{out}} = \frac{\left( R - \frac{R}{4} \right)(R + 4R)}{R^{2}} \Rightarrow B_{out} = \frac{8}{3}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.