Torque (Magnitude of Force) (1 Viewer)

[MrLuvable]

A 5.0cm square coil with 150turns of wire is in a magnetic field of intensity 0.25T. The plane of the coil makes an angle of 30 degrees with the magnetic field. The coil carries a current of 3.0 A.

a) Calculate the magnitude of the force acting on each side of the coil.

I've done the rest of the question, but couldn't get my head around this first bit.

Thanks.

 

[micuzzo]

F=nBILsin@

n=150, I=3.0A B=0.25 L=0.05 @=30

150 x 0.25 x 3.0 x 0.05 x sin30 = 2.8125... is this frm dot point

 

[M@ster P]

no this is from jacaranda question 20

 

[MrLuvable]

It's also in the Dot Point book, thanks

 

[micuzzo]

It's also in the Dot Point book, thanks

yes it looks familar... noworries

 

[helper]

You need to think about the angle between the wire and the magnetic field. The angle remains ninety degrees as the coil rotates, so the force remains the same.

 

[Aerath]

Yeah, I thought it would be sin90? =\

 

[Nuendo]

would the answer be 3.75x10^-2 N? or 5.625 N?

...or neither? lol

 

[Aerath]

Answer is 5.625N. Because @ = 90, therefore sin90 = 1.