The distance between the wire and the circular current loop is r = 2.2 cm . The

Question

The distance between the wire and the circular current loop is r = 2.2 cm .

The distance between the wire and the circular current loop is r = 2.2 cm. (Figure 1) What is the magnitude of the torque on the circular current loop? Express your answer with the appropriate units. Part B What is the loop's stable equilibrium position? The dipole will be in equilibrium after rotating clockwise 180n The dipole will be in equilibrium after rotating clockwise 90r'. The dipole will be in equilibrium after rotating clockwise 135degree The dipole will be in equilibrium after rotating clockwise 45.

Explanation / Answer

a)

here

Iw = 2A

I = 0.2 A

R = 1 * 10^-3 m

r = 0.022 m

first

u = I * A = I * pie * R^2

u = 0.2 * pie * ( 1 * 10^-3)^2 = 6.28 * 10^-7 Am^2

and then

B = u0 * Iw / (2*pie * r)

B = 4 * pie * 10^-7 * 2 / ( 2 * pie * 0.022)

B = 1.81 * 10^-5 T

so the torque is

T = u * B

T = 6.28 * 10^-7 * 1.81 * 10^-5

T = 1.136 * 10^-11 Nm

b)

T = u * B * sin(theta)

for equilibrium orientation the theta will be 0 or 180 deg

so the correct option is 1

the dipole will be in equilibrium after rotating clockwise 180deg

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