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rohit rathi Grade: 12
        

A long thin walled pipe of radius R carries current I along its length.The current density is uniform over circumference of pipe.The magnetic field at the centre of pipe due to quarter portion of pipe?-Please answer and explain.....

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

total current through circumfrence  = I


 current through unit segment of circumfrenece = I/2piR                   (current dencity)


length of quater of segment = 2piR/4 = piR/2


 current through quater segment (piR/2)= (I/2piR)(piR/2)


                                                        =I/4


now we have , B.dl = uoI(current threading)


                    B(2PiR) = uo(I/4)


                    B = uoI/8piR

6 years ago
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