# Work Done in Changing Shape of Current Carrying Loop

• I
Anmoldeep
How would you go about calculating the work done in morphing a square current-carrying loop into a circular current-carrying loop, without change in length while maintaining the same angular orientation with an external magnetic field.

My book suggests defining P(potential energy) = M.B (dot product of magnetic moment and magnetic field)

I am familiar with the above formula for a varying angle between M and B but not for a varying magnetic moment. If it's true please help me in deriving it.
Suppose for the question
1.) Edge if Square loop is 'a'
2.) Current = I
3.) Magnetic field (Uniform and perpendicular to the plane of the loop)

ergospherical
The relevant factor is the ratio ##4/\pi## of the areas Anmoldeep
The relevant factor is the ratio ##4/\pi## of the areas Thanks for that, although I wanted a deeper explanation as to why potential energy is still defined by the same way, P=M.B is found by integrating Torque =MxB w.r.t (theta), here though theta remains the same and M is changing, I have found an integral by mapping every elemental length on the square back to the circle and calculating the work done for all such elemental length's but the integral is not computable, it has truncated trigonometric terms like cos(pi/4tan(theta)-theta) that too under the square root along with other terms.

Anmoldeep
Thanks for that, although I wanted a deeper explanation as to why potential energy is still defined by the same way, P=M.B is found by integrating Torque =MxB w.r.t (theta), here though theta remains the same and M is changing, I have found an integral by mapping every elemental length on the square back to the circle and calculating the work done for all such elemental length's but the integral is not computable, it has truncated trigonometric terms like cos(pi/4tan(theta)-theta) that too under the square root along with other terms.
I think I found the solution, the textbook directly suggested the use of PE=-M.B, however we know that
Del(PE)=M.B(1-cos(theta))=PEtheta - PE0
consider two separate loops of magnetic moment M1 and M2 (square and circle respectively)
Del(PE1)=M1.B(1-cos(theta))=PEtheta - PE0
and
Del(PE2)=M2.B(1-cos(theta))=PEtheta - PE0

choose theta = pi/2 and the PEtheta term will cancel out, subtract the remaining expressions and you get the required answer as the difference in potential energies of the loops when angle between M and B is 0

Last edited:
ergospherical
Thanks for that, although I wanted a deeper explanation as to why potential energy is still defined by the same way, P=M.B is found by integrating Torque =MxB w.r.t (theta)
Recall that the moment ##\mathbf{K}## acting on the magnetic dipole is ##\mathbf{K} = \mathbf{m} \times \mathbf{B}##. If the dipole is rotated to angle ##\theta## (e.g. about an axis perpendicular to the plane containing ##\mathbf{m}## and ##\mathbf{B}##) then the work done by the magnetic field is \begin{align*}
w(\theta) = \int^{\theta} \mathbf{K} \cdot d\boldsymbol{\varphi} = -\int^{\theta} mB \sin{\varphi} d\varphi = mB \cos{\varphi} \bigg{|}^{\theta} = mB\cos{\theta} + c_0
\end{align*}##c_0## can be set arbitrarily. The potential energy is simply ##u = -w = -\mathbf{m} \cdot \mathbf{B}##.

• Anmoldeep
Homework Helper
Gold Member
Supplementary to the above posts, maybe combining them (?):

Ratio of mag moment = ratio of areas for same current.
Then compute ## \Delta PE ## as I think @ergospherical describes.
Then ## \Delta W = \Delta P.E. ##, W = work.