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MCQEasyJEE 2026Ampere's Law

JEE Physics 2026 Question with Solution

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 end of the conductor [B.] maximum at the axis of the conductor and minimum at the midpoint [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

    B, C Only

  • C

    A, D Only

  • D

    E Only

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A long cylindrical conductor of radius RR carries current II uniformly distributed over its cross-section.

Find: Which statement about the magnetic field inside the conductor is correct.

Concept: For a long straight conductor with uniform current density, the magnetic field inside the conductor at distance rr from the axis is

B(r)=μ0Ir2πR2B(r)=\frac{\mu_0 I r}{2\pi R^2}

So,

B∝rB \propto r

This means the magnetic field increases linearly with distance from the axis.

At the axis of the conductor, r=0r=0, hence

B=0B=0

So the magnetic field is minimum at the axis.

At the surface, r=Rr=R, the magnetic field is maximum.

Now verify the statements:

  • [A.] False, because the maximum is not at the ends.
  • [B.] False.
  • [C.] False.
  • [D.] True.
  • [E.] False.

Therefore, only statement D is correct. The correct option is A.

Common mistakes

  • Assuming the magnetic field is maximum at the axis is incorrect because inside a uniformly current-carrying conductor B∝rB \propto r. At r=0r=0, the field is zero, so the axis gives the minimum value, not the maximum.

  • Confusing the field inside the conductor with the field outside the conductor leads to errors. Inside, the enclosed current depends on rr, so the field increases linearly up to the surface.

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