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An ideal solenoid having 40 turns/cm has an aluminium core and carries a current of 2.0 A. Calculate the magnetization l developed in the core and the magnetic field B at the centre. The susceptibility X of aluminimum = 2.3 xx 10^(-5). |
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Answer» SOLUTION :The magnetic intensity H at the centre of the solenoid is H = ni = 4000 turns/m `xx 2.0 A= 8000 A//m` the MAGNETIZATION is `I=etaH` `=2.3xx10^(-5)xx8000A//m=0.18A//m` The magnetic field is `B=mu_(0)(H+I)` `=(4pixx10^(-7)T-m//A)[800+0.18]A//m=3.2pixx10^(-4)T` Note that H >>I in case of a paramagnetic core |
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