GATE 2025 EC Q.65: solution and answer

GATE 2025 · ECQ.65 · 2 marks · NAT

Two resistors are connected in a circuit loop of area $5 m^{2}$ , as shown in the figure below. The circuit loop is placed on the $x$ - $y$ plane.

When a time-varying magnetic flux, with flux-density $B (t) = 0.5 t$ (in Tesla), is applied along the positive $z$ -axis, the magnitude of current $I$ (in Amperes, rounded off to two decimal places) in the loop is ________.

GATE 2025 EC question 65: A single rectangular conducting loop (hatched interior labelled 'Loop Area', with an arrow pointing into the hatched… — Figure from the GATE 2025 EC paper. It is part of the question.

Answer and solution

ANSWER

$∣ I ∣ = \frac{2.5 V}{4 Ω} = 0.625 A \approx 0.63 A$ .

Official answer: 0.62 to 0.63Source: Official answer key, GATE 2025 EC (IIT Roorkee)

Flux through the loop. The field $B = 0.5 t \overset{a}{^}_{z} T$ is uniform and the loop lies in the $x$ - $y$ plane, so $B$ is normal to the loop over its whole area $A = 5 m^{2}$ : $Φ (t) = B (t) A = 0.5 t \times 5 = 2.5 t Wb$
Faraday's law, $E = - \frac{d Φ}{d t}$ : $∣ E ∣ = \frac{d}{d t} (2.5 t) = 2.5 V$ The emf does not change with time.
Loop resistance. Both $2 Ω$ resistors sit in the same single loop, and the connecting wires are ideal, so the resistors are in series: $R = 2 + 2 = 4 Ω$ .
Ohm's law around the loop: $∣ I ∣ = \frac{∣ E ∣}{R} = \frac{2.5}{4} = 0.625 A \approx 0.63 A$
Direction (Lenz's law, for completeness). With $x$ to the right and $y$ up, the right-handed $z$ -axis points out of the page. The flux along $+ \overset{a}{^}_{z}$ is increasing, so the induced current sets up a field along $- \overset{a}{^}_{z}$ (into the page). Seen from $+ z$ , that current circulates clockwise, so it runs left-to-right along the top wire. That is opposite to the arrow marked $I$ , which is why the question asks only for the magnitude.

KEY IDEA

A uniform flux density that rises linearly with time gives a constant motional-free (transformer) emf, $∣ E ∣ = A \frac{d B}{d t}$ . Every element in a single closed loop carries the same current, so the emf drives that current through the total series resistance.

WATCH OUT FOR

Treating the two $2 Ω$ resistors as parallel because they are drawn on opposite sides, which gives $R = 1 Ω$ and $I = 2.5 A$ . There is only one loop, so they are in series. A second slip is using $B = 0.5 t$ itself as the flux and forgetting to multiply by the $5 m^{2}$ area, which gives $0.125 A$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 5 Maxwell's equations questions, with notes

GATE 2025 EC Q.65: solution and answer

Answer and solution

More Maxwell's equations questions