JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

Two identical circular loops P and Q each of radius $r$ are lying in parallel planes such that they have common axis. The current through P and Q are $I$ and $4 I$ respectively in clockwise direction as seen from O. The net magnetic field at O is:

A
$\frac{3 μ _{0} I}{4 2 r}$ toward P
B
$\frac{μ _{0} I}{4 2 r}$ toward P
C
$\frac{μ _{0} I}{4 2 r}$ towards Q
D
$\frac{3 μ _{0} I}{4 2 r}$ towards Q

Solution

With both currents appearing clockwise as seen from O, the magnetic field contributions from the two loops at O are directed away from each loop (along the axis), i.e. in opposite directions at O. The net field points toward the loop with the larger current - toward Q.

For a circular loop of radius $r$ carrying current, the axial field at distance $r$ from its centre is

$B = \frac{μ _{0} I r ^{2}}{2 ( r ^{2} + r ^{2} ) ^{3/2}} = \frac{μ _{0} I}{4 2 r}$ .

$B_{n e t} = \frac{4 μ _{0} I}{4 2 r} - \frac{μ _{0} I}{4 2 r} = \frac{3 μ _{0} I}{4 2 r}$ , directed toward Q.

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