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A long solenoid has 500 turns. When a current of 2A is passed through it., the resultant magnetcis flux linked with each turn of the solenoid is 4xx10^-3Wb. What is the self inductance of the solenoid? |
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Answer» SOLUTION :The total MAGNETIC flux(`phi`) LINKED with the SOLENOID is given by `phi= 500xx4xx10^-3 = 2Wb ` But of a coil `phi = Li ` ` therefore L = phi/i = 2Wb/2A = 1H` |
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