JEE Main 2026 Jan 23 Shift 2, Physics Q4: Magnetic Field due to Electric Current

JEE Main2026Jan 23, Shift 2Physics

The current passing through a conducting loop in the form of equilateral triangle of side $43$ cm is $2$ A. The magnetic field at its centroid is $α \times 1 0^{- 5}$ T. The value of $α$ is ______.

(Given : $μ_{o} = 4 π \times 1 0^{- 7}$ SI units)

A
$23$
B
$3$
C
$33$
D
$\frac{3}{2}$

Solution

Distance from centroid to each side of an equilateral triangle of side $a$ is $d = \frac{a}{2 3}$ . For $a = 43$ cm $= 43 \times 1 0^{- 2}$ m:

$d = \frac{4 3 \times 1 0 ^{- 2}}{2 3} = 2 \times 1 0^{- 2}$ m.

Magnetic field due to one finite side at the centroid (subtending $6 0^{\circ}$ on each side):

$B_{1} = \frac{μ _{0} I}{4 π d} (sin 6 0^{\circ} + sin 6 0^{\circ}) = \frac{μ _{0} I}{4 π d} \cdot 3$ .

By symmetry, total field $B = 3 B_{1} = \frac{3 3 μ _{0} I}{4 π d}$ .

$B = \frac{3 3 \cdot 1 0 ^{- 7} \cdot 2}{2 \times 1 0 ^{- 2}} = 33 \times 1 0^{- 5}$ T.

$∴ α = 33$ .

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JEE Main 2026 Jan 23 Shift 2, Physics Q4: Magnetic Field due to Electric Current

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