In the figure below, a long circular pipe with outside radius R = 2.94 cm carrie
ID: 1461739 • Letter: I
Question
In the figure below, a long circular pipe with outside radius R = 2.94 cm carries a (uniformly distributed) current i = 12.0 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 3.67, but it has the opposite direction.Explanation / Answer
Let the current in wire = iwire
The Magnetic field at the center of the pipe is due to the wire alone, with a magnitude of =
B1 = (uo * iwire) /(2*pi *3R)
B1 = (uo * iwire) / (6*pi*R)
The Magnetic field at point P is due both to the pipe and the wire and is given by -
B2 = (uo * ipipe) / (4*pi*R) - (uo * iwire) / (2*pi*R)
Now, B2/B1 = 3.67
B2 = 3.67 * B1
(uo * ipipe) / (4*pi*R) - (uo * iwire) / (2*pi*R) = 3.67 * (uo * iwire) / (6*pi*R)
ipipe / 4 - iwire/2 = 3.67 * iwire / 6
ipipe / 2 = 3.67 * iwire / 3 + Iwire
6.0 mA = 3.67 * iwire / 3 + Iwire
Iwire = 2.69 mA
(a) Magnitude of current , Iwire = 2.69 mA
(b)
Since the direction of these two is opposite the current in the wire must create an opposite direction field from the pipe at point P and hence it must also be into the page.
Direction is Into the Page.