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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.....

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.....

Grade:12

2 Answers

vikas askiitian expert
509 Points
10 years ago

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

Jerin
19 Points
3 years ago
I current - 2πdI - dΦdI=IdΦ/2πdB=u0dI/2πR =u0IdΦ/4ππRB=integration of dB =u0I[i+j]/4ππRR =u0I√2/4ππRR

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