A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a unifo - Physics MCQ AIEEE 2026 | ExamTest.live

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A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field B going into the plane of the paper. The three currents I_{1}, I_{2} and I (where I is through the connector) are:

Solution & Step-by-step Explanation

The moving connector PQ acts as a battery with emf ε = B v l .The loop consists of two parallel branches, each with resistance R .The equivalent resistance of the two parallel branches is R_{p a r a ll e l} = \frac{R \times R}{R + R} = \frac{R}{2} .The total resistance of the circuit including the connector itself is R_{t o t a l} = R + \frac{R}{2} = \frac{3 R}{2} .Total current I = \frac{ε}{R _{t o t a l}} = \frac{B v l}{3 R /2} = \frac{2 B v l}{3 R} .Since the two parallel branches have equal resistance, the current splits equally: I_{1} = I_{2} = \frac{I}{2} = \frac{B v l}{3 R} .

Try it yourself before checking the explanation above.

A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field B going into the plane of the paper. The three currents I_{1}, I_{2} and I (where I is through the connector) are:

I_{1} = I_{2} = \frac{B v l}{R}, I = \frac{2 B v l}{R}

I_{1} = I_{2} = \frac{B v l}{3 R}, I = \frac{2 B v l}{3 R}

I_{1} = I_{2} = I = \frac{B v l}{R}

I_{1} = I_{2} = \frac{B v l}{6 R}, I = \frac{B v l}{3 R}

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