(8c29p48) A long circular pipe with outside radius R carries a (uniformly distri

Question

(8c29p48) A long circular pipe with outside radius R carries a (uniformly distributed) current i = 10 A into the page as shown in the figure. A wire runs parallel to the pipe at a distance 3R from center to center. Find the magnitude and direction of the current in the wire such that the resultant magnetic field at point P between them and a distance R from the wire has the same magnitude as the resultant field at the center of the pipe but is in the opposite direction. Take positive current as into the page and negative as out of the page.

Explanation / Answer

B field at center of pipe is only from current in wire. Call this current "x". Then we know

B field at center of pipe = k x / 3R

And

B field at point P due to current in pipe is B1 = k i / 2R and this is UP

B field at P from wire is stronger than at the center of the pipe, so we know the B field from wire at P must go against the B field from pipe. That means B field from wire must be DOWN, so current in wire must be INTO THE PAGE.

Now...

B field from wire at P B2 = k x / R

And since the B field at the center of pipe is DOWN (because wire current is into the page), we know the total B field at P is up.

So NOW...

mag (in up direction) of B field at P = magnitude (in down direction) of B field at center of pipe

k i / 2R - k x / R = k x / 3R

Eliminate k and R

i / 2 - x = x / 3

or

10 / 2 = (4/3) x

x = 3.75 A is the current in the wire (Positive because it is into the page). Note that if you plug this back in, you get

B field at P = k * 10 / 2R - k *3.75 / R = 1.25 k/R

and at the center of the pipe

B = k * 3.75 / 3R = 1.25 k/R

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