A circular loop of wire having a radius of 6.0 cm carries a current of 0.14 A .

Question

A circular loop of wire having a radius of 6.0 cm carries a current of 0.14 A. A unit vector parallel to the dipole moment of the loop is given by 0.40i -0.92j. If the loop is located in a magnetic field given by B = (0.40 T)i + (0.35 T)k,

A) find the magnitude of the magnetic dipole moment of the loop.

B) Find the i component of the torque on the loop.

C) Find the j component of the torque on the loop.

D) Find the k component of the torque on the loop.

E) Find the magnetic potential energy of the loop.

Explanation / Answer

A)
m = I*S =0.14*pi*0.06^2 =1.583*10^-3 A*m^2

B, C, D)
Written as vector we have
m = 1.583*10^-3*(0.4*i -0.92*j) =6.33*10^-4*j -1.46*10^-3*j        (A*m^2)
The Torque as vector is
T = m x B = i                                   j                                   k
                     6.33*10^-4               -1.46*10^-3                  0
                     0.40                             0                                   0.35
= i*(-0.511*10^-3) +j*(-2.216*10^-4) +k*(0.584*10^-3)

Tx =-0.511*10^-3 N*m
Ty =-2.216*10^-4 N*m
Tz =+0.584*10^-3 N*m

E) magnetic potential energy is
U =-m*B =-(6.33*10^-4*i -1.46*10^-3*i) *(0.4*i +0.35*k) =-6.33*10^-4*0.4 =-2.532*10^-4 J

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