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A square loop of side `10 cm` and resistance `0.5 Omega` is placed vertically in the east-west plane. A uniform magnetic field of `0.10 T` is set up across the plane in the north-east direction. The magnetic field is decreased to zero in `0.70 s` at a steady rate. The magnetude of current in this time-interval is. |
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Answer» The initial magnetic field is given by `phi = BA cos theta` Given, `B = 0.10 T`, area of sqare loop `= 10 xx 10 = 100 cm^(2) = 10^(-2)m^(2)` `:. phi = (0.1 xx 10^(-2))/(sqrt(2))Wb` Final flux, `phi_("min") = 0` The change in flux is brought about in `0.70 s` The magnitude of the induced emf is `e = (Delta phi)/(Delta t) = (|phi-0|)/(Delta t) = (10^(-3))/(sqrt(2) xx 0.7) = 1mV` The magnitude of current is `I = (e )/(R ) = (10^(-3))/(0.5) = 2mA` |
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