Question

A circular loop of wire having a radius of 8.0 cm carries a current of 0.20 A. A vector of unit length and parallel to the dipole moment W of the loop is given by 0.60 hati-0.80 hatj . if the loop located in uniform magnetic field given by B= (0.25 T)hati+ (0.30 T)hatk find, (a) the torque on the loop and (b) the magnetic potential energy of the loop.

Answer 1

Given:

A circular loop of wire , radius = 8.0 cm

current , I = 0.20 A.

Unit vector along dipole moment W = 0.60 hati-0.80 hatj .

Uniform magnetic field , B= (0.25 T)hati+ (0.30 T)hatk.

To Find:

(a) the torque on the loop

(b) the magnetic potential energy of the loop.

Solution:

Magnetic dipole moment |W| = Current x Area

• |W| = I x πr² = 0.2 x π x 0.08² = 2xπx64 x $10^{-5}$ = 4  x $10^{-3}$ Am²
• W is in the direction of Wu = 0.6i - 0.8j
• W = |W|xWu  = 2.4i - 3.2j x $10^{-3}$ Am²

(a) Torque on the loop,

• T = W x B
• T = $\left[\begin{array}{ccc}i&j&k\\2.4&-3.2&0\\0.25&0&0.30\end{array}\right]$ = i ( -0.96) - j ( 0.72 ) + k (  - 0.8 )

• T = -0.96i -0.72j -0.8k x $10^{-3}$

(b) the magnetic potential energy of the loop.

• U = - M.B

• U = - ( 2.4i - 3.2j x $10^{-3}$)  . ( 0.25 i+ 0.30k )
• U = -0.6 x $10^{-3}$ J

The torque on the loop is  T = -0.96i -0.72j -0.8k x $10^{-3}$ and

the magnetic potential energy of the loop is  U = -0.6 x $10^{-3}$ J.