JEE Main 2025 Jan 28 Shift 2, Physics Q16: Magnetic Field due to Electric Current

JEE Main2025Jan 28, Shift 2Physics

An infinite wire has a circular bend of radius $a$ , and carrying a current $I$ as shown in figure. The magnitude of magnetic field at the origin $O$ of the arc is given by :

A
$\frac{μ _{0} I}{4 π a} [\frac{π}{2} + 1]$
B
$\frac{μ _{0} I}{4 π a} [\frac{3 π}{2} + 1]$
C
$\frac{μ _{0} I}{2 π a} [\frac{π}{2} + 2]$
D
$\frac{μ _{0} I}{4 π a} [\frac{3 π}{2} + 2]$

Solution

The configuration has three parts. Two semi-infinite straight wires whose extensions pass through $O$ contribute, with one ending at the arc's start and the other beginning at the arc's end. The straight portion (1) which has $O$ off its line contributes $\frac{μ _{0} I}{4 π a}$ at perpendicular distance $a$ ; the other one (3) whose line passes through $O$ contributes zero.

The circular arc subtends an angle $θ = 3 π /2$ at the centre $O$ , so the arc's contribution is

$B_{arc} = \frac{μ _{0} I}{4 π a} \cdot θ = \frac{μ _{0} I}{4 π a} \cdot \frac{3 π}{2}$ .

Total $B = \frac{μ _{0} I}{4 π a} [\frac{3 π}{2} + 1]$

Concept behind this question

Magnetic Field due to Electric Current Notes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 28 Shift 2, Physics Q16: Magnetic Field due to Electric Current

Concept behind this question

More previous year questions on Magnetic Field due to Electric Current

Practice more Physics