JEE Main 2026 Jan 28 Shift 2, Physics Q14: Ampere's Circuital Law

JEE Main2026Jan 28, Shift 2Physics

A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is :

A. maximum at either ends of the conductor and minimum at the midpoint

B. maximum at the axis of the conductor

C. minimum at the surface of the conductor

D. minimum at the axis of the conductor

E. same at all points in the cross-section of the conductor

Choose the correct answer from the options given below :

A
D Only
B
A, D Only
C
B, C Only
D
E Only

Solution

For a solid cylinder carrying uniform current, by Ampere's law: inside ( $r < R$ ), $B = \frac{μ _{0} I r}{2 π R ^{2}}$ — linear in $r$ ; outside ( $r > R$ ), $B = \frac{μ _{0} I}{2 π r}$ — falls as $1/ r$ .

$B$ is therefore zero at the axis (minimum) and peaks at the surface ( $r = R$ ). Only statement D is correct.

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JEE Main 2026 Jan 28 Shift 2, Physics Q14: Ampere's Circuital Law

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