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A steady current `i` flows in a small square lopp of wire of side `L` in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let `vecmu_(1)` and `vecmu_(2)` respectively denote the magnetic moments due to the current loop before and after folding. ThenA. `vec(mu_(2))=0`B. `vec(mu_(1))` and `vec(mu_(2))` are in the same directionC. `|vec(mu_(1))|//|vec(mu_(2))|=sqrt(2)`D. `|vec(mu_(1))|//|vec(mu_(2))|=(1//sqrt(2))` |
Answer» Correct Answer - C `|vec(mu)_(1)|=IA,|vec(mu)_(2)|=I(sqrt(2)A)` |
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