JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

The magnetic field at the centre of a current carrying circular loop of radius $R$ is 16 $μ$ T. The magnetic field at a distance $x = 3 R$ on its axis from the centre is ______ $μ$ T.

A
$22$
B
4
C
2
D
8

Solution

Centre: $B_{c} = \frac{μ _{0} I}{2 R} = 16 μ$ T.

On-axis at distance $x$ : $B_{x} = \frac{μ _{0} I R ^{2}}{2 ( x ^{2} + R ^{2} ) ^{3/2}}$ .

With $x = 3 R$ : $x^{2} + R^{2} = 4 R^{2}$ , so $(x^{2} + R^{2})^{3/2} = 8 R^{3}$ .

$B_{x} = \frac{μ _{0} I R ^{2}}{2 \cdot 8 R ^{3}} = \frac{μ _{0} I}{16 R} = \frac{1}{8} B_{c} = \frac{16}{8} = 2 μ$ T.

Option (3).

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →