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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2026 · 28 January · Evening Shift · Physics · Magnetic Effects of Current

A long cylindrical conductor with large cross section carries an...

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 :

Options

  1. AA, D Only
  2. BD Only
  3. CE Only
  4. DB, C Only

Correct answer

B. D Only

Step-by-step solution

For a long cylindrical conductor with uniform current distribution, apply Ampere's law to find the magnetic field.
At distance rr from the axis (inside the conductor, r<Rr < R): B(r)=μ0Ir2πR2B(r) = \frac{\mu_0 I r}{2\pi R^2} (proportional to rr)
At the axis (r=0r = 0): B=0B = 0 (minimum)
At the surface (r=Rr = R): B=μ0I2πRB = \frac{\mu_0 I}{2\pi R} (maximum)
Analyzing the statements:
A. Maximum at ends and minimum at midpoint - FALSE. This describes behavior along the axis, but the magnetic field depends on radial distance, not axial position.
B. Maximum at the axis of the conductor - FALSE. The field is zero at the axis.
C. Minimum at the surface of the conductor - FALSE. The field is maximum at the surface.
D. Minimum at the axis of the conductor - TRUE. The field is zero at r=0r = 0, making it the minimum.
E. Same at all points in the cross-section - FALSE. The field varies from 0 at the axis to maximum at the surface.
Only D is correct.
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