Start free

Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 24 January, Evening Shift — Question 51

. A long straight wire of a circular cross-section with radius 'a' carries a steady current I. The current I is a uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by (1)

  1. Option A:
    Option A figure
    Correct
  2. Option B:
    Option B figure
  3. Option C:
    Option C figure
  4. Option D:
    Option D figure

Answer: A

Step-by-step solution

Bin=(μ0(¯(""J)ׯr))/2B"in"∝rB"out"=(μ0I)/2πrBnet∝1/rB_in=(μ_0 (¯(" " J)ׯr))/2 B_"in " ∝r B_"out " =(μ_0 I)/2πr B_net∝1/r

images

Solution figure

Answer key and solution verified before publishing.

Practise Moving Charges and Magnetic Field

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law

More Moving Charges and Magnetic Field questions from this paper

. A long straight wire of a circular cross-section with radius 'a'… | JEE Main 2025 PYQ with Solution · DhiX AI