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The magnetic flux near the axis inside a current-carrying air-core solenoid is pi/3 xx 10^(-6) Wb. What is its magnetic moment if the length of the solenoid is 60 cm? [Assume the length to be large compared with the cross section of the solenoid] |
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Answer» Solution :The magnetic INDUCTION near the AXIS, and well inside, of an air-core solenoid is `B=(Phi_(m)/A) = mu_(0)(N/L)I`, where N is the number of turns. Therefore, its magnetic moment, `M=NIA = (Phi_(m)L)/(mu_(0)) = ((pi/3 xx 10^(-6)) xx 0.6)/(4pi xx 10^(-7)) = 0.5 A-m^(2)` |
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