In the figure 29-71, a long circular pipe with outside radius R = 1.55 cm carrie

Question

In the figure 29-71, a long circular pipe with outside radiusR = 1.55 cm carries a (uniformly distributed) currenti = 11.0 mA into the page. A wire runs parallel to thepipe at a distance of 3.00R from center to center. Findthe magnitude of the current in the wire in milliamperes such thatthe ratio of the magnitude of the net magnetic field at pointP to the magnitude of the net magnetic field at the centerof the pipe is 4.39, but it has the opposite direction.

Just confused about this one and what to do. Thanks!

Explanation / Answer

The pipe, due to symmetry,does not contribute to BC. In other words, the B fieldat the center of the pipe due to the current in the pipe is zero.So BC must be entirely from the current in thewire. 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 from the pipe       Bpipe = o I2 / 2 (2R) BP is the sum of these. i.e, Bp= o I1 / 2 R+o I2 / 2 (2R) And since BP  =- BC ,then -Bc=o I1 / 2 R+ oI2 / 2 (2R)                       -o I1 / 2 (3R) =o I1 / 2 R+ oI2 / 2 (2R)       o I1 /2 3R    +    oI1 / 2 R   + oI2 / 2 2R    = 0 2I1   +   6I1    + 3I2     = 0 I1 = -(3/8)I2        means I1 is out of the page.

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