B_axis = μ₀IR²/(2(R²+x²)^(3/2)). At center B₀ = μ₀I/(2R). For B/8: solving gives x = √3R
Back to JEE Main 2023 Electromagnetism
JEE Main 2023 — Physics Electromagnetism
hard
mcq
2023
Official previous-year question
Verified 30 May 2026.
Question
A current carrying circular loop of radius R produces a magnetic field B at its center. At what distance from center on its axis, the magnetic field becomes B/8?
Options
- AR
- B√3R
- C2R
- DR/√3
Solution
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Electromagnetism in other years
More JEE Main Electromagnetism Questions
A charged particle moves in a uniform magnetic field. If the velocity is perpendicular to the field, the path of the particle is:
2026
easy
mcq
A long straight wire carries a current of 10 A. The magnetic field at a distance of 0.1 m from the wire is:
2026
easy
mcq
The magnitude of force on a charge q moving with velocity v in a magnetic field B at angle θ is:
2025
medium
mcq
The electrostatic force between two point charges q₁ and q₂ separated by distance r is given by Coulomb's law as:
2024
easy
mcq
The electric field at a distance r from an infinitely long uniformly charged wire having linear charge density λ is:
2023
medium
mcq