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

The figure shows a long, coaxial cable in which a current $I$ flows down through the inner cylinder of radius $a$ and the same current flows back through the outer cylinder of radius $2 a$ . The cylinders are insulated from each other and the current is uniformly distributed over the area of the cross-section in each cylinder. The strength of the magnetic field at a distance $3 a / 2$ from the axis of the cable is $B = \frac{7 μ_{0} I}{n π a}$ . The value of $n$ is___________

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

answer is 36.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Using Ampere's circuital law for a closed loop of radius $r = \frac{3 a}{2}$ round the axis of the cable.
$\oint_{} \vec{B} \cdot \overset{⇀}{d l} = μ_{0} I_{n e t}$

$B [2 π (\frac{3 a}{2})] = μ_{0} [I - \frac{I}{π [{(2 a)}^{2} - a^{2}]} \times π [{(\frac{3 a}{2})}^{2} - a^{2}]]$

$B [3 π a] = μ_{0} I [1 - \frac{1}{3} \times [\frac{9}{4} - 1]]$

$B [3 π a] = μ_{0} I [1 - \frac{5}{12}]$

$B = \frac{7 μ_{0} I}{36 π a}$

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!