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A right-angled isosceles triangular conducting loop of height 10 cm is positioned so that its 90 deg vertex almost touches an infinitely long straight current-carrying wire, with the loop's hypotenuse running parallel to the wire (the wire is insulated from the loop). A counterclockwise current in the loop is increased steadily at 10 A/s. Which statement(s) is/are correct?

1. The current induced in the long wire flows opposite to the loop current along the hypotenuse.
2. The wire and the loop repel each other.
3. Rotating the loop about the wire at constant angular speed induces an extra emf of (μ0/π) volt in the wire.
4. The induced emf magnitude in the wire is (μ0/π) volt.

Correct answer: The induced emf magnitude in the wire is (μ0/π) volt.

Solution

This is a multiple-correct problem; the cleanly computable numeric result is the magnitude of the mutually induced emf. By Faraday/mutual-inductance reciprocity, the emf appearing across the wire due to dI/dt in the loop is M*(dI/dt). Carrying out the flux integral over the isosceles right triangle with dI/dt = 10 A/s gives exactly (μ0/π) volt. The induced current in the wire opposes the change, and an increasing loop current also produces a repulsive interaction, so statements A and B are physically correct as well; rotating about the wire does not add a steady extra emf, so C is false.

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