GATE 2026 EE Q.51: solution and answer

GATE 2026 · EEQ.51 · 2 marks · MCQ

The figure shows an arbitrarily shaped planar conducting loop A in the XY plane. Two nonintersecting regions with areas $a_{1}$ and $a_{2}$ within the loop are subjected to magnetic fields $B_{1} = \frac{m}{2} sin (ω t) (1 \overset{x}{^} + 0 \overset{y}{^} + 1 \overset{z}{^})$ , and $B_{2} = - \frac{n}{2} cos (2 ω t + π /4) (0 \overset{x}{^} + 1 \overset{y}{^} + 1 \overset{z}{^})$ , respectively.

What is the expression for the induced rms voltage in loop A?

GATE 2026 EE question 51: A 3-D axis sketch with Z pointing up, Y pointing right and X pointing down-left toward the viewer, all meeting at the… — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$\frac{a _{1}^{2} ω ^{2} m ^{2} + 4 a _{2}^{2} ω ^{2} n ^{2}}{4}$
(B)
$\frac{a _{1}^{2} ω ^{2} m ^{2} + 4 a _{2}^{2} ω ^{2} n ^{2}}{2}$
(C)
$\frac{a _{1}^{2} ω ^{2} m ^{2} - 2 a _{2}^{2} ω ^{2} n ^{2}}{2}$
(D)
$a_{1}^{2} ω^{2} m^{2} + 2 a_{2}^{2} ω^{2} n^{2}$

Answer and solution

ANSWER

The induced rms voltage is $\frac{a _{1}^{2} ω ^{2} m ^{2} + 4 a _{2}^{2} ω ^{2} n ^{2}}{4}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2026 EE (IIT Guwahati)

The loop lies in the XY plane, so its area vector is along $\overset{z}{^}$ (the sign depends on the direction of traversal). Only the $z$ -component of each field links flux: $Φ = \int B \cdot d S = \int B_{z} d S$ . The $\overset{x}{^}$ and $\overset{y}{^}$ components lie in the plane of the loop, so they link no flux.
Take each field as uniform over its patch. The $z$ -components are $B_{1 z} = \frac{m}{2} sin (ω t)$ and $B_{2 z} = - \frac{n}{2} cos (2 ω t + π /4)$ . The fluxes are therefore $Φ_{1} = a_{1} \frac{m}{2} sin (ω t)$ and $Φ_{2} = - a_{2} \frac{n}{2} cos (2 ω t + π /4)$ .
By Faraday's law, $e = - \frac{d Φ}{d t}$ with $Φ = Φ_{1} + Φ_{2}$ . Since $\frac{d Φ _{2}}{d t} = \frac{2 a _{2} nω}{2} sin (2 ω t + π /4)$ ,
$e (t) = - \frac{a _{1} mω}{2} cos (ω t) - \frac{2 a _{2} nω}{2} sin (2 ω t + π /4)$
The two terms are sinusoids at different frequencies, $ω$ and $2 ω$ . Their peak values are $E_{1} = \frac{a _{1} mω}{2}$ and $E_{2} = \frac{2 a _{2} nω}{2}$ .
Over the common period $2 π / ω$ , the average of the cross product $cos (ω t) sin (2 ω t + π /4)$ is zero, because sinusoids at different frequencies are orthogonal. The mean-square values therefore add: $E_{r m s}^{2} = \frac{E _{1}^{2}}{2} + \frac{E _{2}^{2}}{2}$ .
Substituting the peaks:
$E_{r m s}^{2} = \frac{1}{2} \cdot \frac{a _{1}^{2} m ^{2} ω ^{2}}{2} + \frac{1}{2} \cdot \frac{4 a _{2}^{2} n ^{2} ω ^{2}}{2} = \frac{a _{1}^{2} ω ^{2} m ^{2} + 4 a _{2}^{2} ω ^{2} n ^{2}}{4}$
Therefore $E_{r m s} = \frac{a _{1}^{2} ω ^{2} m ^{2} + 4 a _{2}^{2} ω ^{2} n ^{2}}{4}$ , which is option (A). The direction of traversal only fixes the overall sign of $e (t)$ , so it does not affect the rms value.

KEY IDEA

A loop in the XY plane links flux only through $B_{z}$ . Faraday's law turns each patch's flux into a sinusoidal emf. Differentiating $cos (2 ω t + π /4)$ doubles the amplitude of the second term. Because the two emfs are at different frequencies ( $ω$ and $2 ω$ ), their rms values combine as $E_{r m s} = E_{1, r m s}^{2} + E_{2, r m s}^{2}$ , not by adding peaks or phasors.

WATCH OUT FOR

Using the full field magnitude, including the in-plane $\overset{x}{^}$ or $\overset{y}{^}$ component, instead of only $B_{z}$ . Other mistakes: forgetting the factor $2$ that the derivative of $cos (2 ω t + π /4)$ brings out, which gives option (D)'s $2 a_{2}^{2}$ coefficient; forgetting to divide the squared peaks by 2 when converting to rms, which leads to option (B); and treating the two emfs as phasors at one frequency and adding them.

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GATE 2026 EE Q.51: solution and answer

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