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JEE Advanced 2026 · Paper 1, Physics question 15

List-I contains four conducting loops lying in the XYXY plane, as shown in the figures. The loops are rotating about ZZ axis passing through the point OO with time period TT in clockwise direction. The region x>0x>0 contains a uniform magnetic field BB in the +z+z direction. List-II contains the qualitative variation of the induced current i(t)i(t) for each of these loops. Choose the option which describes the correct match between the entries in List-I to those in List-II.

List-I

  • P
    A semicircular loop in x < 0 with its diameter along the y-axis, centred at O; the field B (out of the page) fills x > 0. The loop rotates clockwise about O.
  • Q
    A loop in x < 0 made of two 60° sectors (above and below O, with a notch between them) whose straight edge lies on the y-axis; field in x > 0; rotating clockwise about O.
  • R
    A single 60° sector loop with its vertex at O and one edge along the +y axis, lying in x < 0; field in x > 0; rotating clockwise about O.
  • S
    Two 60° sectors with vertices at O on opposite sides: one in x > 0 (upper right) and one in x < 0 (lower left), joined at O; field in x > 0; rotating clockwise about O.

List-II

  • 1
    i against t: a positive pulse for a short time just after t = 0, zero until T/2, a negative pulse just after T/2, then zero.
  • 2
    i against t: two positive pulses in the first half period and two negative pulses in the second half.
  • 3
    i against t: constant positive current for the first half period, then constant negative current of the same size for the second half.
  • 4
    i against t: zero for the whole period.
  • 5
    i against t: current rising linearly from zero over the first half period, then negative and growing linearly in magnitude in the second half.
Choose one option
Answer and worked solution

Official answer: C

The emf is −dΦdt-\dfrac{d\Phi}{dt} and Φ=B×(loop area in x>0)\Phi=B\times(\text{loop area in }x>0). Rotating clockwise, each part of a loop enters the field across the +y+y axis and leaves across the −y-y axis. A sector of angle ϕ\phi sweeps area at a constant rate while it crosses, taking time ϕ2πT\frac{\phi}{2\pi}T.

(P) The semicircle enters steadily for T/2T/2, then leaves steadily for T/2T/2: constant current, reversing at T/2T/2 → 3.

(R) One 60∘60^\circ sector, starting at the +y+y axis: it enters during 00-T/6T/6 and leaves during T/2T/2-2T/32T/3. One positive pulse, then one negative pulse → 1.

(Q) Sectors at 90∘90^\circ-150∘150^\circ and 210∘210^\circ-270∘270^\circ.

  • Upper sector: enters 00-T/6T/6, leaves T/2T/2-2T/32T/3.
  • Lower sector: enters T/3T/3-T/2T/2, leaves 5T/65T/6-TT.

Two positive pulses in the first half, two negative in the second → 2.

(S) The two sectors are diametrically opposite. As one leaves the field the other enters at the same rate, so the area in the field never changes: no current → 4.

P→3, Q→2, R→1, S→4.

Topics:Faraday's and Lenz's lawsMotional EMF
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