JEE Main 2025 Apr 3 Shift 1, Physics Q24: Magnetic Field due to Electric Current

JEE Main2025Apr 3, Shift 1Physics

A loop ABCDA, carrying current $I = 12$ A, is placed in a plane, consists of two semi-circular segments of radius $R_{1} = 6 π$ m and $R_{2} = 4 π$ m. The magnitude of the resultant magnetic field at center O is $k \times 1 0^{- 7}$ T. The value of $k$ is _______ . (Given $μ_{0} = 4 π \times 1 0^{- 7}$ Tm A $^{- 1}$ )

Your answer

Solution

Straight segments AB and CD pass through O along the line of the wire, so each contributes zero field at O.

The two semicircles are concentric at O. Field at centre of a semicircle of radius $R$ carrying current $I$ :

$B = \frac{μ _{0} I}{4 R} .$

Given the current direction, the two semicircular fields are antiparallel at O, so they subtract:

$B_{O} = \frac{μ _{0} I}{4 R _{2}} - \frac{μ _{0} I}{4 R _{1}} = \frac{μ _{0} I}{4} (\frac{1}{4 π} - \frac{1}{6 π}) .$

$= \frac{4 π \times 1 0 ^{- 7} \times 12}{4} \cdot \frac{1}{π} (\frac{1}{4} - \frac{1}{6}) = 12 \times 1 0^{- 7} \cdot \frac{1}{12} = 1 \times 1 0^{- 7} T .$

$∴ k = 1.$

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JEE Main 2025 Apr 3 Shift 1, Physics Q24: Magnetic Field due to Electric Current

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