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JEE Main 3 August 2021, Shift 1, Physics Q12

Question 12 of 90 in this shift, Physics question 12 of 30, Section A.

Consider the case of a non-homogeneous solid cylinder of radius R carrying a current ‘I' such that the current density depends on the radial distance ‘

r

' from the axis of the cylinder as J =

\sigma r

, (

\sigma

is constant). The magnetic field at a point ‘P' at a perpendicular distance

r

(< R) from the axis of the cylinder is -

(1) $\frac{\mu_0 \mathrm{I} \mathrm{R}^2}{2\pi r^3}$
(2) $\frac{\mu_0 \mathrm{I} r^2}{2\pi \mathrm{R}^3}$ Official answer
(3) $\frac{\mu_0 \mathrm{I} r^2}{2\pi \mathrm{R}^2}$
(4) $\frac{\mu_0 \mathrm{I} \mathrm{R}}{2\pi r^2}$

Official answer

Option 2

NTA final key.

Chapter: Moving Charges and Magnetism
Topic: Ampere's Circuital Law
Idea tested: Non-Uniform Current Density in Cylinders

Same topic in other shifts

All Ampere's Circuital Law questions

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.