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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.....

8 years ago

Answers : (2)

vikas askiitian expert
509 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

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