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JEE Advanced 2022 · Paper 2, Physics question 18

Which one of the following options represents the magnetic field B⃗\vec B at O due to the current flowing in the given wire segments lying on the xyxy plane?

Wire carrying current I in the xy plane, with point O marked. A straight vertical wire of length L carries I upward; its top end is level with O. From its top, a horizontal segment of length L/2 runs right to the left end of a semicircular arc of radius L/2 centred at O, which goes over the top (clockwise) to its right end; a dashed line marks the diameter through O. A quarter-circle arc of radius L/4 centred at O (labelled L/4, with a dotted radius from O) runs clockwise from the dashed line on the right of O down to the point L/4 directly below O. From there a straight vertical wire of length 3L/4 carries I downward. Axes drawn alongside: i to the right, j up, k out of the plane (drawn toward the lower left).
Choose one option
Answer and worked solution

Official answer: C

Segments that contribute nothing. The horizontal segment and the lower vertical wire both lie on lines through O, so they give no field there. So does the short radial piece along the dashed line.

The arcs (field into the plane, −k^-\hat k, for the clockwise current):

  • Semicircle, radius L2\frac L2: μ0I4(L/2)=μ0I2L\dfrac{\mu_0I}{4(L/2)}=\dfrac{\mu_0I}{2L}.
  • Quarter circle, radius L4\frac L4: μ0I8(L/4)=μ0I2L\dfrac{\mu_0I}{8(L/4)}=\dfrac{\mu_0I}{2L}.

The long vertical wire. It is at distance LL from O and runs from O's level down a length LL, so it subtends angles from 00 to 45∘45^\circ:
B=μ0I4πLsin⁡45∘=μ0I42πL,also along −k^.B=\frac{\mu_0I}{4\pi L}\sin45^\circ=\frac{\mu_0I}{4\sqrt2\pi L},\quad\text{also along }-\hat k.

Total.
B⃗=−μ0IL(1+142π)k^.\vec B=-\frac{\mu_0I}L\left(1+\frac1{4\sqrt2\pi}\right)\hat k.
(C).

Topics:Biot-Savart law
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