In the figure 29-71, a long circular pipe with outside radius R = 2.6 cm carries

Question

In the figure 29-71, a long circular pipe with outside radiusR = 2.6 cm carries a (uniformly distributed) currenti = 8.00 mA into the page. A wire runs parallel to thepipe at a distance of 3.00R from center to center. Findthe (a) magnitude and (b) direction (into or out of the page) ofthe current in the wire such that the net magnetic field at pointP has the same magnitude as the net magnetic field at thecenter of the pipe but is in the oppsite direction. PLEASE HELP ME. THIS IS DUE TODAY T_T LIFESAVE FOR WHOEVERANSWER THIS QUESTION!!!

Explanation / Answer

Given : BC = -BP    where the minus sign indicatesopposite direction BC + BP  =   0 The pipe, due to symmetry, does not contribute toBC. In other words, the B field at the center of thepipe due to the current in the pipe is zero. So BC mustbe entirely from the current in the wire. BC = o I1 /2 3R Now the field at point P is due to both the current inthe wire and the current in the pipe. The contribution from thewire is:        Bwire = o I1 / 2 R and Bcylinder = o I2 / 2 2R BP is the sum of these. And sinceBC and BP sum to zero we can add all thesetogether     o I1 /2 3R    +    oI1 / 2 R   + oI2 / 2 2R    = 0 2I1   +   6I1    + 3I2     = 0 I1 = -(3/8)I2      you knowI2 = 8.0 mA    I1 = -3mA means I1 is out of the page. I hope it helps you BC + BP  =   0 The pipe, due to symmetry, does not contribute toBC. In other words, the B field at the center of thepipe due to the current in the pipe is zero. So BC mustbe entirely from the current in the wire. BC = o I1 /2 3R Now the field at point P is due to both the current inthe wire and the current in the pipe. The contribution from thewire is:        Bwire = o I1 / 2 R Now the field at point P is due to both the current inthe wire and the current in the pipe. The contribution from thewire is:        Bwire = o I1 / 2 R and Bcylinder = o I2 / 2 2R BP is the sum of these. And sinceBC and BP sum to zero we can add all thesetogether     o I1 /2 3R    +    oI1 / 2 R   + oI2 / 2 2R    = 0 2I1   +   6I1    + 3I2     = 0 I1 = -(3/8)I2      you knowI2 = 8.0 mA    I1 = -3mA means I1 is out of the page. I hope it helps you I hope it helps you

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