# What is the magnitude of the torque acting on the loop? T = IAB

1. Nov 2, 2005

### mr_coffee

A circular wire loop of radius 20.0 cm carries a current of 3.10 A. It is placed so that the normal to its plane makes an angle of 34.0° with a uniform magnetic field of 15.0 T.

I don't understand why this isnt right, i got the first part right, it wanted me to find the magnetic dipole moment of the loop, which i found to be .3896 Am^2; I got this by using M = IA;
now it wants me to find the magnitude of the torque acting on the loop. So I used the formula:
Torque = IAB;
I = current
A = area of loop
B = Magnetic field;
Torque = (3.10)(PI*.20^2)(15) = 5.84 Nm, which is wrong, any ideas why?
Thanks.

2. Nov 2, 2005

### James R

The torque is only IAB if the field lines are perpendicular to the plane of the loop, whioch isn't the case here.

The complete formula is:

$$\tau = IAB \sin \theta$$

where $\theta$ is the angle between the normal to the loop and the magnetic field. In your problem, the relevant angle is 34 degrees.

In vector terms, the torque is:

$$\vec{\tau} = \vec{\mu} \times \vec{B}$$

where $\vec{\mu}$ is the magnetic moment.

3. Nov 2, 2005

### mr_coffee

ohh thanks alot james, worked great!