JEE Advanced 2022 Paper 2 Q18 Physics Magnetism Biot-Savart Law Medium

JEE Advanced 2022 Paper 2 · Q18 · Biot-Savart Law

Which one of the following options represents the magnetic field $\vec{B}$ at $O$ due to the current $I$ flowing in the given wire segments lying on the $xy$ plane?

[Figure: A wire carrying current I in the xy-plane composed of segments: two horizontal segments of length L/2 meeting at O at the top, two slanted segments of length L/4 each meeting at O at 45 degrees, a vertical segment of length L on the left, and a horizontal segment of length 3L/4 at the bottom. Coordinate axes (i, j, k) shown with k out of the page.]

  1. A. $\vec{B} = -\dfrac{\mu_0 I}{L}\left(\dfrac{3}{2} + \dfrac{1}{4\sqrt{2}}\right)\hat{k}$
  2. B. $\vec{B} = -\dfrac{\mu_0 I}{L}\left(\dfrac{3}{2} + \dfrac{1}{2\sqrt{2}}\right)\hat{k}$
  3. C. $\vec{B} = -\dfrac{\mu_0 I}{L}\left(1 + \dfrac{1}{4\sqrt{2}}\right)\hat{k}$
  4. D. $\vec{B} = -\dfrac{\mu_0 I}{L}\left(1 + \dfrac{1}{4}\right)\hat{k}$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Decompose the wire into segments and apply Biot-Savart for finite straight wires: $B = (\mu_0 I)/(4\pi r)(\sin\alpha_1 + \sin\alpha_2)$, with $r$ the perpendicular distance from $O$ to the segment. Segments along which $O$ lies on the line of the wire (collinear segments) contribute zero: this kills $B_{AB}, B_{DE}, B_{FG}$ in the official decomposition. Remaining contributions (each into the page, $-\hat{k}$): - Two slant segments at $45^\circ$ each at perpendicular distance from $O$: $B = \mu_0 I/(4\pi L) \sin 45^\circ$ (using geometry yields the $1/(4\sqrt{2})$ term). - Long straight segment with $O$ on its perpendicular bisector contributing $\mu_0 I/(\pi L \cdot 4)$. Summing as in the FIITJEE working: $\vec{B} = -(\mu_0 I/L)\,(1/(4\sqrt{2}) + 1)\,\hat{k} = -(\mu_0 I/L)(1 + 1/(4\sqrt{2}))\hat{k}$. Official key: (C).

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