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Physics
Classical Physics
Electromagnetism
Work Done in Changing Shape of Current Carrying Loop
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[QUOTE="Anmoldeep, post: 6532264, member: 693572"] 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))=PE[SUB]theta[/SUB] - PE[SUB]0[/SUB] consider two separate loops of magnetic moment M[SUB]1[/SUB] and M[SUB]2[/SUB] (square and circle respectively) Del(PE[SUB]1[/SUB])=M[SUB]1[/SUB].B(1-cos(theta))=PE[SUB]theta[/SUB] - PE[SUB]0[/SUB] and Del(PE[SUB]2[/SUB])=M[SUB]2[/SUB].B(1-cos(theta))=PE[SUB]theta[/SUB] - PE[SUB]0[/SUB] choose theta = pi/2 and the PE[SUB]theta[/SUB] 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 [/QUOTE]
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Physics
Classical Physics
Electromagnetism
Work Done in Changing Shape of Current Carrying Loop
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