GATE 2022 IN Q.57: solution and answer

GATE 2022 · INQ.57 · 2 marks · NAT

A conducting semi-circular loop of radius $R = 0.1 m$ , with its diameter centered at the origin, rotates in the $x$ - $y$ plane about the origin with a constant angular velocity, $ω = 20 rad/s$ , as shown. A magnetic field of magnitude $B = 2 T$ and normal to $x$ - $y$ plane exists in the region $x \geq 0$ as shown. If the loop has a resistance of $2 Ω$ , and negligible inductance, the peak-to-peak current (in milliamperes) in the loop is ________ (round off to one decimal place)

GATE 2022 IN question 57: Standard x-y axes with the origin marked (0,0) by a dot. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

The peak-to-peak current is 200.0 mA. The current is a square wave that switches between +100 mA and -100 mA every half-turn.

Official answer: 200.0Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Flux linked. Only the part of the half-disc lying in $x \geq 0$ links flux. The diameter is centred at the origin, so the half-disc is a sector of angle $π$ with its vertex at the origin. The field region $x \geq 0$ is a half-plane bounded by a line through that same origin, so it is also a sector of angle $π$ about the origin. Two sectors with the same vertex overlap in a sector, whose angle $α (t)$ is the overlap of their angular ranges. So $Φ (t) = B \cdot \frac{1}{2} R^{2} α (t)$
How the overlap changes. The half-disc spans an angle of $π$ , and so does the field half-plane. As the loop turns at $ω$ , the overlap angle grows linearly from $0$ to $π$ over half a turn, then falls linearly back to $0$ over the next half turn. Its rate is always the rotation rate: $\frac{d α}{d t} = \pm ω$ . So $Φ (t)$ is a triangular wave in time, not a sinusoid.
Faraday's law. $∣ ε ∣ = \frac{d Φ}{d t} = \frac{1}{2} B R^{2} ω = \frac{1}{2} (2) (0.1)^{2} (20) = 0.2 V$ The magnitude is the same at every instant. Only its sign changes, at the moments when $α$ reaches $0$ or $π$ .
Ohm's law (inductance is negligible, so the current follows the emf at once): $∣ i ∣ = \frac{∣ ε ∣}{R _{loop}} = \frac{0.2}{2} = 0.1 A = 100 mA$
Lenz's law gives the sign. While the overlap grows (more flux into the page), the induced current makes a field out of the page, so it flows counter-clockwise as drawn. While the overlap shrinks, it flows clockwise. So $i (t)$ is a square wave that switches between $+ 100 mA$ and $- 100 mA$ every half-turn, i.e. every $π / ω \approx 0.157 s$ .
Peak-to-peak current: $i_{pp} = 100 - (- 100) = 200.0 mA$

KEY IDEA

The half-disc and the field region $x \geq 0$ are both sectors of angle $π$ centred on the pivot, so their overlap is always a sector. Its area $\frac{1}{2} R^{2} α$ rises and falls linearly at the rate $ω$ . That makes the flux a triangular wave, so the emf has a constant magnitude $\frac{1}{2} B R^{2} ω$ and flips sign every half-turn. The current is a $\pm 100 mA$ square wave, and its peak-to-peak value is twice the magnitude.

WATCH OUT FOR

There are three common mistakes. (1) Treating this like a coil turning in a uniform field and assuming a sinusoidal emf. Here the flux is a triangular wave and the emf is a square wave. (2) Reporting the magnitude, 100 mA, and forgetting that the current reverses when the overlap starts to shrink. The peak-to-peak value is 200 mA. (3) Deciding the current is DC because the emf magnitude never changes, and answering 0. It is also easy to use the whole semicircle's area $π R^{2} /2$ in place of the rate of change of the overlap sector, $\frac{1}{2} R^{2} ω$ .

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 5 Maxwell's equations questions, with notes

GATE 2022 IN Q.57: solution and answer

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