In the figure, part of a long insulated wire carrying current i = 5.89 mA is ben

Question

In the figure, part of a long insulated wire carrying current i = 5.89 mA is bent into a circular section of radius R = 1.37 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

Given

i = 5.89 *10^-3 A

R = 1.37*10^-2 m

the magnetic field at the center of curvature due to straight section and due to circular section is  

say B1,B2

due to straight section, B at c is B1= mue0*i/2pi*R

and due to circular section B at C is B2 = mue0*i/2R

the net field at C , B = B1+B2

B = (mue0*i/2R)(1+1/pi)

B = ((4pi*10^-7*5.89*10^-3)/(2*1.37*10^-2))(1+1/pi) T

B = 3.5611651*10^-7 T

by symmetry we have the resultant magnetic field is along z direction so  

Bx= 0 T , By = 0 t, Bz = 3.5611651*10^-7 T

b)

now the loop is perpendicular to the plane of the page

then

B1 and B2 are perpendicular to each other that the angle between them is 90 degrees

From Parallelogram law of vectors the magnitude of the resulatant vector is  

B = sqrt(B1^2+B2^2+2B1*B2 cos90)

B = srt(B1^2+B2^2)

B = sqrt((mue0*i/2pi*R)^2+(mue0*i/2R)^2)

B = (mue0*i/2*R)sqrt(1+1/pi^2)

B = (4pi*10^-7*5.89*10^-3/(2*1.37*10^-2)) sqrt(1+1/2pi) T

B = 2.9083411*10^-7 T

the angle between the fields is theta = arc tan (B1/B2)

theta= arc tan ((mue0*i/2pi*R)/(mue0*i/2R))

theta = arc tan (1/pi) = 17.66 degrees

so the components of the field along  

X is Bx = B cos theta = 2.908341*10^-7 cos 17.66 T

Bx = 2.771281*10^-7 T

along y is

By = B sin theta = 2.908341*10^-7 sin 17.66 T = 8.822973*10^-8 T

along z

Bz = 0

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