help with QAT 09 question 14 - electric power

[TyErd]

help.

[shokstar]

This has confused the crap out of me as well..

For a, i thought the force on a side, F=bil, so the force would always be the same no matter what orientation...?

For b, i always thought the force on BC is always 0...?

For c, since the motor presumably travels at the same speed, the turning effect, or the torque, is always constant...?

For d, I thought the net force is also 0 as well, as the force on Ab and DC are opposite, so they cancel out and the net force is 0.....?

[TyErd]

that's exactly what i thought haha.

[ghadz7]

@ least post the question up!

[TyErd]

I've only got the hard copy of it....

[shokstar]

Here. It relates to the diagram on the right of page 1, the question is on the left of page 2.

[TyErd]

a. 1 and 2
b. 2
c. 1
d. 1,2 and 3

[TyErd]

anyone know how to do this?

[Linkage1992]

The only explanation I can think of is assuming diagram 2 is still completely within the magnetic field, in which case the force on side AB would be the same in 1 and 2 as it is perpendicular to the field in both cases.

For BC, I think it's meant to be a compromise. In position 1 it is parallel to the field so there will be no force. In position 3 it is perpendicular but we can assume only a very small length of the side is actually within the magnetic field, meaning the force will be minimized.

In position 2 it has a certain angle to the field, and it is still completely within the field (as assumed) and thus at this point it will have the maximum force.

For the turning effect, it is a maximum when the coil is parallel to the field (diagram 1) and the net force is constant unless external factors are changed. This last part, however, I need clarification on, as I'm unsure of the effect of the coil not being completely within the field in 3 on the net force...