JEE Main 2026 · 2 Apr · Shift 1 · Physics Q39

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PhysicsSingle correct+4 / -1Question 39 of 75

Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres P and Q of a semicircular arc are $B_1$ and $B_2$ respectively, then the ratio $\dfrac{B_1}{B_2}$ is ________.

(I) A hairpin: two long parallel straight wires (current I to the right in the upper wire, I to the left in the lower wire) joined at their right ends by a semicircle of radius r centred at P. (II) A long straight wire carrying I to the right ends at the top of a semicircle of radius r centred at Q; from the bottom of the semicircle the wire continues straight down (current I downward), the vertical line passing through the end of the arc directly below the horizontal wire's end.

\dfrac{2+\pi}{1+\pi}

\dfrac{1+\pi}{1-\pi}

\dfrac{2+\pi}{1-\pi}

\dfrac{1+\pi}{2-\pi}

Answer and worked solution

Official answer: A

Building blocks. A semicircle gives $\dfrac{\mu_0I}{4r}$ at its centre; a semi-infinite wire ending at the foot of the perpendicular gives $\dfrac{\mu_0I}{4\pi r}$ . All contributions here are along the same direction.

(I) Two straight wires and the arc: $B_1=\dfrac{\mu_0I}{4\pi r}(2+\pi)$ .

(II) The horizontal wire and the arc. The vertical wire's line passes through Q, so it contributes nothing: $B_2=\dfrac{\mu_0I}{4\pi r}(1+\pi)$ .

Ratio. $\dfrac{B_1}{B_2}=\dfrac{2+\pi}{1+\pi}$ .

(A)

Topics:Biot-Savart law

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JEE Main 2026 · 2 Apr · Shift 1, Physics question 39