In the figure 29-71, a long circular pipe with outside radius R = 2.99 cm carrie

Question

In the figure 29-71, a long circular pipe with outside radius R = 2.99 cm carries a (uniformly distributed) current i = 10.1 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the magnitude of the current in the wire in milliamperes such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is 3.91, but it has the opposite direction.
Figure 29-71 step by step please... Thank you

Explanation / Answer

From fig we can write BC = -BP where the minus sign indicates opposite 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 = µo I1 /2p 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 = µo I1 / 2p R and Bcylinder = µoI2 / 2p 2R BP is the sum of these. And sinceBC and BP sum to zero we can add all thesetogether µo I1 /2p 3R + µoI1 / 2p R + µoI2 / 2p 2R = 0 2I1 + 6I1 + 3I2 = 0 I1 = -(3/8)I2 Given that I2 = 10.1 mA ==> I1 = -3.7875 mA negative sign indicates I1 is out of the page.

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