Chapter 29, Problem 020 In the figure, part of a long insulated wire carrying cu

Question

Chapter 29, Problem 020

In the figure, part of a long insulated wire carrying current i = 2.14 mA is bent into a circular section of radius R = 8.11 cm. What are (a) the x-component, (b) the y-component, and (c) the z-component of the magnetic field at the center of curvature C if the circular section lies in the plane of the page as shown? What are (d) the x-component, (e) the y-component, and (f) the z-component of the magnetic field at the center of curvature C if the circular section is perpendicular to the plane of the page after being rotated 90° counterclockwise as indicated?

Explanation / Answer

Magnetic field due to the wire:

B1 = u0*i/(2*pi*R)

Magnetic field due to the loop

B2 = u0*i/(2*R)

if the circular section lies in the plane of the page as shown, then

Bnet = B1 + B2

Bnet = [u0*i/(2*R)]*(1 + 1/pi)

Bnet = [4*pi*10^-7*2.14*10^-3/(2*8.11*10^-2)]*(1 + 1/pi)

Bnet = 2.18*10^-8 T

from Right hand rule direction of magnetic field will be out of the page in z direction, So

A.

x-component = 0 T

B.

y-component = 0 T

C.

z-component = 2.18*10^-8 T

Now if if the circular section is perpendicular to the plane of the page after being rotated 90° counterclockwise as indicated, then

Bnet = sqrt (B1^2 + B2^2)

Bnet = [u0*i/(2*R)]*sqrt (1 + (1/pi)^2)

Bnet = [4*pi*10^-7*2.14*10^-3/(2*8.11*10^-2)]*sqrt (1 + (1/pi)^2)

Bnet = 1.74*10^-8 T

Angle of Bnet will be

theta = arctan (B1/B2) = arctan (1/pi) = 17.65 deg

D.

Bx = 1.74*10^-8*cos 17.65 deg = 1.66*10^-8 T

E.

By = 0 T

F.

Bz = 1.74*10^-8*sin 17.65 deg = 0.527*10^-8 T

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