AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A solid cylindrical conductor of radius $R_{1}$ is kept inside a hollow cylindrical conductor of inner radius $R_{2}$ and outer radius $R_{3}$ $(R_{1} < R_{2} < R_{3})$ in a co-axial manner, as shown. The solid cylinder carries a current $l_{1}$ and the hollow conductor carries current $l_{2}$ in the opposite directions. The two conductors are kept insulated from each other and current through each is distributed uniformly over the cross-section. Let $r$ be any non-zero radial distance from the axis.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Magnitude of magnetic induction B at points inside the solid conductor, so that $r < R_{1}$ , can be zero

Magnitude of magnetic induction B at points inside the solid conductor, so that $r < R_{1}$ , can be $\frac{μ_{0} I_{1}}{2 π R_{1}^{2}} r$

Magnitude of magnetic induction B at points outside the solid conductor such that $R_{1} < r < R_{2}$ will be $\frac{μ_{0} (I_{1} - I_{2})}{2 π r}$

For $R_{2} < r < R_{3}$ , magnitude of B can be expressed as $\frac{μ_{0}}{2 π r} [I_{1} - \frac{I_{2} (r^{2} - R_{2}^{2})}{(R_{3}^{2} - R_{2}^{2})}]$

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$a) r < R_{1} :$ By Ampere's circuital law, $B = \frac{μ_{0} I_{1} r}{2 π R_{1}^{2}}$

Option (A) is wrong and Option (B) is correct

$c) R_{1} < r < R_{2} :$ By Ampere's circuital law, $B = \frac{μ_{0} I_{1}}{2 π r}$

Option (C) is wrong

$d) R_{2} < r < R_{3} :$ $\int \vec{B} \cdot \vec{d l} = μ_{0} [i_{n e t}]$

$B .2 π r = μ_{0} [I_{1} - I_{2}^{1}]$

$B = \frac{μ_{0}}{2 π r} [I_{1} - \frac{I_{2}}{π (R_{3}^{2} - R_{2}^{2})} . π (r^{2} - R_{2}^{2})]$

$B = \frac{μ_{0}}{2 π r} [I_{1} - \frac{I_{2} (r^{2} - R_{2}^{2})}{(R_{3}^{2} - R_{2}^{2})}]$

Option (D) is correct

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!