JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres $P$ and $Q$ of a semicircular arc are $B_{1}$ and $B_{2}$ respectively, then the ratio $\frac{B _{1}}{B _{2}}$ is ________.

A
$\frac{2 + π}{1 + π}$
B
$\frac{1 + π}{1 - π}$
C
$\frac{2 + π}{1 - π}$
D
$\frac{1 + π}{2 - π}$

Solution

Field from an infinite straight wire at perpendicular distance $r$ is $\frac{μ _{0} I}{2 π r}$ , and from a semicircular arc of radius $r$ at its centre is $\frac{μ _{0} I}{4 r}$ .

Configuration I (centre $P$ ): two collinear semi-infinite straight wires contribute the same field as a single infinite straight wire (both at distance $r$ from $P$ , with consistent direction at $P$ because of the bump geometry):

$B_{1} = \frac{μ _{0} I}{4 π r} \times 2 + \frac{μ _{0} I}{4 r} = \frac{μ _{0} I}{4 r} (\frac{2}{π} + 1) = \frac{μ _{0} I ( 2 + π )}{4 π r}$

Configuration II (centre $Q$ ): one semi-infinite straight wire plus the semicircular arc:

$B_{2} = \frac{μ _{0} I}{4 π r} + \frac{μ _{0} I}{4 r} = \frac{μ _{0} I}{4 r} (\frac{1}{π} + 1) = \frac{μ _{0} I ( 1 + π )}{4 π r}$

$∴ \frac{B _{1}}{B _{2}} = \frac{2 + π}{1 + π}$

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