JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :

A
$\frac{μ _{0}}{2 π} \frac{I}{r} (π + 1) \hat{i}$
B
$- \frac{μ _{0}}{2 π} \frac{I}{r} (π - 1) \hat{i}$
C
$\frac{μ _{0}}{2 π} \frac{I}{r} (π - 1) \hat{i}$
D
$- \frac{μ _{0}}{2 π} \frac{I}{r} (π + 1) \hat{i}$

Solution

Three contributions at the centre O:

(i) The two straight semi-infinite parts (AB and DE) end at the points where the circle meets the line, each a perpendicular distance $r$ from O. Each contributes $\frac{μ _{0} I}{4 π r}$ in the $+ \hat{i}$ direction (their fields add).

(ii) The full circular loop at its centre contributes $\frac{μ _{0} I}{2 r}$ in the $- \hat{i}$ direction (opposite sense from the straight parts based on the current direction shown).

Net: $B_{0} = \frac{μ _{0} I}{4 π r} \hat{i} + \frac{μ _{0} I}{4 π r} \hat{i} - \frac{μ _{0} I}{2 r} \hat{i} = \frac{μ _{0} I}{2 π r} \hat{i} - \frac{μ _{0} I}{2 r} \hat{i} = \frac{μ _{0} I}{2 π r} (1 - π) \hat{i} = - \frac{μ _{0} I}{2 π r} (π - 1) \hat{i}$ .

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