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Question

42. A 100 millihenry coil carries a current 1A .Energy stored in its magnetic field is
43. If the number of turns per unit length of a coil of solenoid is reduced to half, the self inductance of the
solenoid will

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Solution

Dear Student ,
42. Energy stored in the magnetic field is ,
E = 1/2 LI2 = 0.05 J
43. As we know that from the equation self inductance L is directly proportional to the square of the number of turns per unit length then we can write that ,when the number of turns per unit length of the coils halved then the self inductance of the solenoid will become 1/4 times .
Regards

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