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A solenoid is of length 50 cm and has a radius of 2cm. It has 500 turns. Around its central section a coil of 50 turns is wound. Calculate the mutual inductance of the system. |
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Answer» <P> Solution :`N_(P)=500, N_(S)=50, A=pi XX 0.02 xx 0.02 m^(2)``mu_(0)=4pi xx 10^(-7) H m^(-1), I=50 cm=0.5 m` 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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