Find time dependent torque needed to turn a wire loop in a uniform B field

[JFuld]

Homework Statement

a single loop of wire with radius A is rotating in a uniform magnetic field at angular frequency ω. The loop of wire drives a current through a resistor R. Determine the torque as a function of time needed to turn the wire.

Homework Equations

1) Emf = ∫Fmag.dl (integral on closed loop)

2) flux of B = $\Phi$ =∫B.da

3) Emf= - d$\Phi$/dt

4) V=IR

The Attempt at a Solution

Here is my procedure right now:

Fist find the time dependent flux of B through the loop.

flux= ∫B.da;

B.da = B*da*cos(θ), where theta = ωt

then the integral is ∫∫B*cos(wt)rdrdθ (theta goes from 0-2pi, r goes from 0-R)

=π(A^2)Bcos(wt)

From this, the Emf generated can easily be found from eq 1.

Emf =-( -sin(wt)*(π(A^2)B/w))

= sin(wt)*(π(A^2)B/w)

Then time dependent current (I) can be found. since Emf=V=IR, I(t)=Emf/R

I(t) = 1/R* sin(wt)*(π(A^2)B/w)


Im not really sure what to do next, but here is what i did:


so the magnetic force per unit charge (Fmag) depends on the velocity of the charges.

The velocity has two parts, velocity due to the angular motion (Ill designate as v) and velocity due to the current (designated as u).


v points in an angular direction (same as direction of rotation)
u points tangent to the wire circle (same direction as current)

to add to the confusion, the current (and thus u) changes direction every half rotation.

then Fmag = (v x B) + (u x B)

letting B be oriented in the +y direction, (v x B) points in the direction of the current

The direction of (u x B) is tricky, it is in the direction normal to both I and B, and has an angular component.

So to find the torque, I need to figure out how (u x B) changes with time. Since this component of Fmag is the only one with an angular component, it is this anguar component that will determine the torque.

Does this sound right? If so, any tricks on figuring out (u x B)?

 

[BruceW]

From this, the Emf generated can easily be found from eq 1.

Emf =-( -sin(wt)*(π(A^2)B/w))

why do you have 1/w ? I don't think you did the differentiation correctly.

Then time dependent current (I) can be found. since Emf=V=IR, I(t)=Emf/R

I(t) = 1/R* sin(wt)*(π(A^2)B/w)

This is right (apart from the 1/w). From here, I guess there are a couple of ways you could go. There is one easy but cheap method that I can think of. You know that energy is being lost by the resistor, so can you use an assumption to give yourself an easy calculation?