Solenoid

Asked by  | 22nd Feb, 2010, 09:38: AM

Expert Answer:

Dear Student

Magnetic field inside the solinoid  B= mu 0    NI/L= 4π  X10-7X(300X3)I/0.60.=I  X  18.8X10-4 -------------4

 

To lift the wire F=BIL=mg= 2.5X10-3 X 10=2.5X10-2  N---(1)

BIL = B 6X0.02=0.12B-----------------2

1=2   ==>      0.12B=2.5X10-2   ==>  B=0.208-----------3

3,4 ==>  I  X  18.8X10-4=0.208===>  I=110.6A 

So ans wer is b

 

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Answered by  | 22nd Feb, 2010, 06:29: PM

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