Ncert Exercises & Solutions

Find the current in the wire for the configuration shown in the figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. $θ$ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

Hint: The induced emf depends on the change in magnetic flux.

Step 1: The induced emf is given by;

$ε = |- \frac{dϕ}{dt}| = |- \frac{d (BA)}{dt}| = B \frac{dA}{dt} = Bd \frac{dx}{dt} = vdB$

Step 2: This induced emf make the flow of current in the closed circuit of resistance R.

$I = \frac{dvB}{R}$ and is independent of q.

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