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

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A circular loop of wire having a radius of6.0 cm carries a currentof 0.14 A. A unit vectorparallel 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 + (0.35T)k,

A) find the magnitude of the magnetic dipolemoment of the loop.

B) Find the i componentof the torque on the loop.

C) Find the j componentof the torque on the loop.

D) Find the k componentof the torque on the loop.

E) Find the magneticpotential energy of the loop.

Answer & Explanation Solved by verified expert

4.1 Ratings (579 Votes)

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*jÂ Â Â Â Â Â Â (A*m^2)
The Torque as vector is
T = m x B = iÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â jÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â k
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 6.33*10^-4Â Â Â Â Â Â Â Â Â Â Â Â Â Â -1.46*10^-3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.40Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 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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