# Magnetic dipole

1. Apr 22, 2005

### dibilo

i recieved this problem from my lecturer

A solenoid is wound with 2200 turns per meter, and carries a current 2.5A. Take the axis of the solenoid to the solenoid to be parallel to the z-axis. A magnetic dipole moment given (m)=(20A.m^2)i+ (30A.m^2)k is located inside the solenoid.

i)how much energy would be required to rotate this magnetic dipole until (m) is parallel to the x-axis?

what i did was to find the magnetic dipole using |m|=sprt(20^2+30^2)= 36A.m^2

then i find the B field to be 6.9x10^-3 k

Uinitial = -(m).B= -(mxBx + myBy + mzBz), where by Bx and By =0 due to the solenoid being // to z-axis = -mzBz = -0.21J

Here is where my problem lies......

i wrote
Ufinal = -(20i).(6.9x10^-3) = 0 , due to Bx = 0 and mz = 0

but my tutor wrote on my worksheet that i should be using 36A.m^2 instead of 20, while in the 1st part, when finding Uinitial, i can use mzBz (30 x 6.9x10^3).

can anyone help me out on this? please as far as possible explain as simple as you can why is this so? thank you very much

Last edited: Apr 22, 2005
2. Apr 22, 2005

### OlderDan

I am having some difficulty following this but I think I see it. The final energy is zero, so the "error" is of no consequence, but the final configuration has the dipole aligned with the x-axis, and the strength of the dipole is 36, not 20. You have rotated it to a new orientation. The x-component is now the only component.