In the figure below, a long circular pipe with outside radius R = 2.76 cm carrie

Question

In the figure below, a long circular pipe with outside radius R = 2.76 cm carries a (uniformly distributed) current i = 14.1 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the (a) magnitude and (b) direction (into or out of the page) of the current in the wire 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 4.25, but it has the opposite direction.

Explanation / Answer

magnetic field at P is B1 = mu_o*I1/(2*pi*2R) - (mu_o*I)/(2*pi*R)

mu_o is the permeability of free space

I1 = 14.1*10^-3 A

R = 2.76*10^-2 m

I is the current in wire

then B1 = ((4*3.142*10^-7*14.1*10^-3)/(2*3.142*2*2.76*10^-2))-((4*3.142*10^-7*I)/(2*3.142*2.76*10^-2))

B1 = 5.1*10^-8 - (7.25*10^-6*I)

magnetic field at the center of the pipe is B2 = mu_o*I/(2*pi*3R) =(4*3.142*10^-7*I)/(2*3.142*3*2.76*10^-2) = 2.42*10^-6 T

Given that

B1 / B2 = 4.25

B1 = 4.25*B2

5.1*10^-8 - (7.25*10^-6*I) = 4.25*2.42*10^-6*I

I = 2.9 mA into the page

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