A loop ABCDA, carrying current I = 12 A, is placed in a plane, consists of two semi-circular segments of radius R₁ = 6π m and R₂ = 4π m. The magnitude of the resultant magnetic field at center O is k × 10⁻⁷ T. The value of k is_________. ( Given μ₀ = 4π × 10⁻⁷ T m A⁻¹) Numerical answer type. Work it out, then open the answer.

Answer: 1. Straight parts AB and CD lie along a line through O: no field. Inner arc ( R₂) is clockwise, outer arc ( R₁) is anticlockwise, so the fields are opposite. B = (μ₀ I)/(4)(1/R₂-1/R₁) = 4π×10⁻⁷×124(1/4π-1/6π) B = 12π×10⁻⁷×(1)/(12π) = 1×10⁻⁷ T So k = 1.

A loop ABCDA, carrying current I = 12 A, is placed in a plane… | JEE Main 2025 (3 Apr, Shift 1)

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Biot-Savart Law & Circular Coils

A loop ABCDA, carrying current $I = 12\text{ A}$ , is placed in a plane, consists of two semi-circular segments of radius $R_1 = 6\pi\text{ m}$ and $R_2 = 4\pi\text{ m}$ . The magnitude of the resultant magnetic field at center O is $k \times 10^{-7}\text{ T}$ . The value of k is_________.

( Given $\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 1

Straight parts AB and CD lie along a line through O: no field.
Inner arc ( $R_2$ ) is clockwise, outer arc ( $R_1$ ) is anticlockwise, so the fields are opposite.
$B = \dfrac{\mu_0 I}{4}\left(\dfrac{1}{R_2}-\dfrac{1}{R_1}\right) = \dfrac{4\pi\times10^{-7}\times12}{4}\left(\dfrac{1}{4\pi}-\dfrac{1}{6\pi}\right)$
$B = 12\pi\times10^{-7}\times\dfrac{1}{12\pi} = 1\times10^{-7}$ T
So $k = 1$ .

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