In the figure below, part of a long insulated wire carrying a current of I = 5.1

Question

In the figure below, part of a long insulated wire carrying a current of I = 5.12 mA is bent into a circular section of radius R = 2.04 cm. What is the magnetic field at point C (the center of the circular section) under the following conditions? (a) Orientation A: The circular section lies in the xy plane (as shown in the figure above). B vector = nT i cap + nT j cap+ nT k cap (b) Orientation B: The circular section is given an additional 90 degree twist (around the pivot axis shown) so that it lies in the yz plane. B vector = nT i cap + j cap nT + nT k cap (c) If we treat the straight section of wire separately from the circular section, we can ask how they interact with each other. If the circular section were free to twist from rest, which would be its preferred orientation (relative to the straight section)? Orientation A Orientation B

Explanation / Answer

current, I=5.12 mA

radius, r=2.04 cm

a)

magnetic field due to circular wire,

B1=uo*I/2r

magnetic field due to stright wire,

B2=uo*I/2pi*r

net magnetic field, Bnet=B1+B2

=uo*I/2r + uo*I/2pi*r

=4pi*10^-7*5.12*10^-3/(2*2.04*10^-2) + 4pi*10^-7*5.12*10^-3/(2pi*2.04*10^-2)

=2.08*10^-7 T

===>

Bnet=(0)i + (0)j +(2.08*10^-7)k

b)

Bnet=(uo*I/2r)i + (uo*I/2pi*r)j

= (4pi*10^-7*5.12*10^-3/(2*2.04*10^-2))i + (4pi*10^-7*5.12*10^-3/(2pi*2.04*10^-2))j

=(1.58*10^-7)i + (5.02*10^-8)j + (0)k

c)

orientation A

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