The magnetic field at the centre of a wire loop formed by two semicircular wires of radii R₁ = 2π m and R₂ = 4π m, carrying current I = 4 A as per figure given below is α × 10⁻⁷ T. The value of α is ________. (Centre O is common for all segments) Numerical answer type. Work it out, then open the answer.

Answer: 3. Straight parts lie along a line through O: no field. In this single loop the two semicircles lie on opposite sides of O and are traversed in the same rotational sense, so their fields add. B = (μ₀ I)/(4)(1/R₁+1/R₂) = 4π×10⁻⁷×44(1/2π+1/4π) B = 4π×10⁻⁷×3/4π = 3×10⁻⁷ T So α = 3.

The magnetic field at the centre of a wire loop formed by two… | JEE Main 2024 (27 Jan, Shift 2)

Q43JEE Main 2024 · 27 JAN · SHIFT 2Medium

Biot-Savart Law & Circular Coils

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $R_1 = 2\pi\text{ m}$ and $R_2 = 4\pi\text{ m}$ , carrying current $I = 4\text{ A}$ as per figure given below is $\alpha \times 10^{-7}\text{ T}$ . The value of $\alpha$ is ________. (Centre $O$ is common for all segments)

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

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Answer: 3

Straight parts lie along a line through O: no field.
In this single loop the two semicircles lie on opposite sides of O and are traversed in the same rotational sense, so their fields add.
$B = \dfrac{\mu_0 I}{4}\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right) = \dfrac{4\pi\times10^{-7}\times4}{4}\left(\dfrac{1}{2\pi}+\dfrac{1}{4\pi}\right)$
$B = 4\pi\times10^{-7}\times\dfrac{3}{4\pi} = 3\times10^{-7}$ T
So $\alpha = 3$ .

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