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[ombuds]

At the instant shown the plane of the coil is parallel to the direction of the magnetic field.
The magnetic field strength is 0.45 T. When the current to the coil is activated it has a magnitude of 1.75 A in the direction ADCB.
(a) Calculate the magnitude and direction of the force acting on side CD when the
current is flowing and the coil is in the position shown in the diagram.

 

[Chambalam]

F = Field Strength x Current x Length of Conductor x Angle (perpendicular here, so equal to 1)
F = 1.75 x 0.45 x 0.05 = 0.039375 = 0.04N

Using the right hand push rule... Fingers point to the right, thumb points out of the page, so the palm points upwards.

0.04N upward.

 

[ombuds]

Thats the answer that i got, But the text book has it at 0.98N up

 

[ombuds]

0.04 was one of the answers i got but i am so confused.

 

[Squishxmishyx]

Text book is wrong. He is always right.

 

[ombuds]

I take your word for it.

 

[Chambalam]

Took a look in the textbook.

"A student makes a model motor. She makes a rectangular coil with 25 turns"

If you use the full answer, 0.039375, multiply by 25 (as the length is 25 as long - 25 turns), then you will get your 0.98

 

[ombuds]

I have no idea how i missed that part of the question. Thanks so much for your help.

 

[Chambalam]

No problem, motors and generators is a little iffy :\

 

[ombuds]

Have you finished it yet? Were upto the torque bit and were supposed to have it finished in 3 weeks. Impossible, but my teacher has assured us that we will get it done and he is a senior marker for Physics and lectures at University level so i would assume he knows what he is talking about.