A single-turn rectangular wire loop measures 6.00 cm wide by 10.0 cm long. The l

Question

A single-turn rectangular wire loop measures 6.00 cm wide by 10.0 cm long. The loop carries a current of 6.00 A. The loop is in a uniform magnetic field with 8 6.00 x 10-3 T. Taking torques about an axis, parallel to either side of the rectangular loop, that maximizes the torque, what is the magnitude of the torque exerted by the field on the loop if the direction of the magnetic field is described as the following? (a) parallel to the short sides of the loop N·m (b) parallel to the long sides of the loop N·m (c) perpendicular to the plane of the loop N·m

Explanation / Answer

(a) parallel to the short sides of the loop?
We can figure out the forces experienced by each side using the right hand rule,
to get a good visualization of what’s causing the torque, and what direction it
would be in. However, we can also use the rule:

= IAB sin

where is the angle between the magnetic field and the area vector of the loop.
In this case, B~ and A~ are perpendicular, so:

= IAB = 6.00 A * (0.06 m × 0.1 m) * 6.00 × 103 T = 0.000216Nm

(b) parallel to the long sides of the loop?

We get the same answer here as B~ and A~ are still perpendicular, so = 0.000216 Nm

(c) perpendicular to the plane of the loop?

In this case, B~ and A~ are parallel (because A~ sticks out perpendicular to the

loop), so = 0 Nm

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