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Question

A uniform time-varying magnetic field exists in a circular region of radius $R$, directed perpendicular into the plane of the paper, increasing at a constant rate $\alpha$. A straight conducting rod of length $2R$ is placed exactly along the diameter of the circular region (passing through the centre). Find the induced emf across the rod.

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$\frac{1}{2} \pi R^2 \alpha$

To solve this problem, we must calculate the induced electromotive force (emf) across the conducting rod that is placed along the diameter of a circular region where a uniform time-varying magnetic field exists. Let's break down the solution step-by-step:

  1. The magnetic field is uniform and increases at a constant rate \(\alpha\). This means the magnetic flux through the circular region is also changing at a rate \(\alpha\).
  2. The change in magnetic flux \(\Delta \Phi_B\) through the circle of radius \(R\) is given by the area of the circle times the rate of increase of the magnetic field: 
\[\Delta \Phi_B = \pi R^2 \alpha\]
  1. According to Faraday's law of electromagnetic induction, the induced emf (\(\mathcal{E}\)) is equal to the negative rate of change of magnetic flux through the circuit. However, for calculating magnitude of emf, we can ignore the negative sign: 
\[\mathcal{E} = \left| \frac{d\Phi_B}{dt} \right|\]
  1. For the rod placed on the diameter, the potential difference across its ends due to the changing magnetic field is a result of the average emf induced along its length. Due to symmetry, the induced emf along the rod is half of that through the entire circle: 
\[\mathcal{E} = \frac{1}{2} \pi R^2 \alpha\]
  1. Thus, the correct answer is the option with the induced emf: 
\[\frac{1}{2} \pi R^2 \alpha\]

Therefore, the induced emf across the rod is \(\frac{1}{2} \pi R^2 \alpha\).

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