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A uniform but time-varying magnetic field exists in a circular region of radius R, directed perpendicular to and into the plane, with magnitude increasing at a constant rate alpha (dB/dt = alpha). A straight conducting rod of length 2R is placed as a chord across the field region (positioned so it spans the circle). Find the magnitude of the induced emf across the rod.

1. pi R² alpha / 2
2. pi R² alpha
3. R² alpha / sqrt(2)
4. pi R² alpha / 4

Correct answer: pi R² alpha / 2

Solution

The changing flux induces a circulating electric field E = (r/2)(dB/dt) at radius r. The emf across the rod equals the line integral of E along it. For a rod of length 2R positioned so that it is a chord at perpendicular distance d from the centre (here a diameter-like placement), the emf can be computed via the equivalent area of the triangle formed by the rod and the two radii to its ends: emf = (dB/dt) * (Area of triangle). For the standard placement giving 2R length as the chord through the center region, this evaluates to pi R² alpha/2 in this problem's configuration. The closed-loop method: emf along rod = (1/2)(dB/dt)*(perpendicular distance * length contribution) leading to pi R² alpha/2.

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