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Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^(1). If the cut is joined and the loop has a resistance of 1.6Omega, how much power is dissipated by the loop as heat? What is the source of this power ? |
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Answer» Solution :When loop is stationary, its SURFACE area `A = 1B = 0.08 xx 0.02 = 16 xx 10^(-4) m^(2)` and rate of change of magnetic field `(dB)/dt = - 0.02 Ts^(-1)` (-ve sign means that the field is gradually falling with time) `therefore` Induced EMF `varepsilon =-(dphi_(B))/dt=16xx10^(-4)xx(-0.02)=3.2xx10^(-5)V` and power dissipated as heat `P=varepsilon^(2)/R=(3.2xx10^(-5))^(2)/1.6=6.4xx10^(-10)W.` The SOURCE of power is the external agency due to which magnetic field is being REDUCED with time. |
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