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A long conducting wire `AH` is moved over a conducting triangular wire `CDE` with a constant velocity `v` in a uniform magnetic field `vec(B)` directed into the plane of the paper. Resistance per unit length of each wire is `rho`. Then A. (a) a constant clockwise induced will flow in the closed loopB. (b) an increasing anticlockwise induced current will flow in the closed loopC. (c ) a decreasing anticlockwise induced current will flow in the closed loopD. (d) an constant anticlockwise induced current will flow in the closed loop |
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Answer» Correct Answer - D (d) Magnetic flux in `ox` direction through the coil is increasing. Therefore, induced current will produce magnetic field in `o.` direction. Thus, the current in the loop is anticlockwise. Magnitude of induced current at any instant of time is `I = (e)/(R) = (Bv(FG))/(rho(FG + GD + DF))` When the wire `AH` moves downwards `FG, GD and DF` all increasing in the same ratio. Therfore, `i` is constant. |
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