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A solenoid is of length 50cm and has a radius of 2cm. It has 500 turns. Around its central section of coil of 50 turns is wound. Calculate the mutual inductance of the system. |
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Answer» Solution :`N_(P)= 500, N_(S)= 50` `A= PI xx 0.02 xx 0.02m^(2)` `mu_(0) = 4pi xx 10^(-7) HM^(-1), l= 50CM= 0.5m` Now, `M= (mu_(0)N_(P)N_(S)A)/(l)` `=(4pi xx 10^(-7) xx 500 xx 50 xx pi xx (0.02)^(2))/(0.5)H` `= 789.8 xx 10^(-7)H= 78.98 mu H` |
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