Torque due to electromagnetic induction

[eas123]

Homework Statement

A circular coil of radius r carries a current I. A magnetic induction B acts at right angles to a diameter of the coil. Show that the current experiences a torque T about the diameter given by T=Iπ$r^{2}$Bsinω, where ω is the angle between the normal to the plane of the coil and the induction B.

Homework Equations

F=∫IdlxB

The Attempt at a Solution

T=Fr/2=Ir/2∫dlxB
Taking the modulus of both sides, T=Ir/2∫Bdlcos(ω)=IrBlcos(ω)/2=Iπ$r^{2}$Bcosω
I suspect the step where I take the modulus of the inside of the integral is wrong, but I'm not sure why.

More generally, I'm unsure how to calculate a cross product such as dlxB around a circle. The modulus of this is Bdlsin(θ), but what is the angle θ?

Thanks in advance for any help! :-)

 

[rude man]

It's easier if you let ω = π/2 at first. So draw the circular loop in the x-y plane with center at the origin and radius = R. Draw a diameter along the x axis. What direction is the B field in? Call θ the angle between the x-axis and a point on the loop in the 1st quadrant.

Then use d F= I d l x B to get the force on the loop in the direction you're looking for. Remember, they're asking for torque about the diameter, so you're looking for the component of d F = dF in the appropriate direction.

Now you have dF about the diameter as a function of θ. Now, given dF, what is torque d τ about the diameter? Watch out , this is tricky - the torque is about the diameter, not the origin, so don't just use dτ = R * dF.

Then integrate dτ from θ = 0 to π/2 and multiply by 4 since there are 4 quadrants.

Finally, include the effect of ω not being π/2 but any angle 0 < ω < π/2. This is a tilt angle about the diameter.

 

[eas123]

Thanks so much for your help. I'll try the method you suggested now.

 

[eas123]

Hi. I'm trying to use the method above but I'm having some problems.
dT=rcosθdF, and T=4∫dT between 0 and π/2. But I don't know how to write dF as a function of dθ.
How do I use the expression dF=I dl x B to find dF?

Thanks so much!

 

[rude man]

Don't start figuring torque. Start with force.

Did you draw the diagram I suggested? Let's take two points on the loop, one at (R,0) the other at (0,R). What is dl x B as a function of R and dθ for those two locations? Remember, B points in the j direction if you let ω = π/2 for the time being.

 

[eas123]

This is the point I don't understand. Since dl and B are perpendicular, dl x B = B dl, but that's not a function of theta...

 

[rude man]

This is the point I don't understand. Since dl and B are perpendicular, dl x B = B dl, but that's not a function of theta...

d l and B are not perpendicular at (R,0) which would correspond to θ = 0 if you drew my diagram.