Calculating Magnetic Dipole Moment & Torque of Coil in Magnetic Field

In summary, a circular coil with 20 turns and a radius of 12.8 cm carries a current of 2.70 A and is placed in a uniform magnetic field of 60.0 mT with an initial angle of 26.0o between the coil's magnetic moment and the magnetic field. The magnetic dipole moment of the coil is found to be 2.779 A*m^2 and the torque acting on the coil is 7.31E-2 N*m. To find the work required to rotate the coil so that the B-field lies in the plane of the coil, the equation T=uoBsin(theta) can be used, where uo is the magnetic dipole moment, B
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Homework Statement



A circular coil having 20 turns and a radius of R = 12.8 cm carries a current of 2.70 A. It is placed in a uniform magnetic field of 60.0 mT. The initial angle between the magnetic moment of the coil μ and the magnetic field B is 26.0o.
1)Calculate the magnetic dipole moment of the coil.
2)What is the magnitude of the torque acting on the coil?
3)How much work is required to rotate the coil so that the B-field lies in the plane of the coil?


Homework Equations



x=uo M/B

The Attempt at a Solution


so I solved for uo by using the equation uo=NAI and got 2.779 but now I do not know what to do and if the equation my professor gave is correct. Then on parts 2 and 3 I don't know where to begin.
 
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  • #2
I got part one and part 2. I think I misunderstood the problem because part one is asking for uo which I solved for and got 2.78 A*m^2 which was correct. Then with this I can find the torque for part 2 with the equation T=uoBsin(theta). So I used this equation and I got 7.31E-2 N*m which was correct. Now I have part 3-how much work is needed-to solve for and i am not sure what I am supposed to do.
 

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