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A coil of resistance 100 `Omega` having 100 turns is placed in a magnetic field. A galvanometer of resistance `400 Omega` is connected in series with it. If the coil is brought from the present magnetic field to another magnetic field in `(1)/(10)s`, determine the average emf and the current. Given, the initial and final magnetic flux linked with each turn of the coil are 1 mWb and 0.2 mWb respectively. |
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Answer» Change in magnetic flux for each turn `=0.2-1=-0.8mWb` So, change in magnetic flux for 100 turns `=100xx(-0.8)mWb=-0.08Wb` Hence, the magnetic of average emf induced = the negative of the rate of change of magnetic flux `=-(-(0.08Wb)/((1)/(10)s))=0.8V` The equivalent resistance of the circuit = 100 + 400 = 500 `Omega`. Hence, the average induced current = `(0.8V)/(500Omega)=0.0016A=1.6mA`. |
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