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In figure (a) a solenoid's magnetic field, increasing into the page, induces an emf in a surrounding conducting loop that lights two bulbs A and B (in series around the loop). In figure (b) two points P and Q on the loop are connected by a short wire. After the short is inserted, what happens to the bulbs?

1. Bulb A goes out, bulb B gets brighter
2. Bulb B goes out, bulb A gets brighter
3. Bulb A goes out, bulb B gets dimmer
4. Bulb B goes out, bulb A gets dimmer

Correct answer: Bulb A goes out, bulb B gets brighter

Solution

Adding the short PQ provides a zero-resistance path that bypasses one bulb (A), so no current flows through it and it goes out. The remaining bulb B now carries the loop's induced emf across less total resistance, so it glows brighter. (The exact bulb depends on the figure; with A in the bypassed branch, A goes out and B brightens.)

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