Electrodynamics question - Induced EMF

[mrmoriarty]

> A very long straight wire carries a current I. A plane rectangular coil of high resistance, with sides of length $a$ and $b$, is coplanar with the wire. One of the sides of length $a$ is parallel to the wire and a distance $D$ from it; the opposite side is further from the wire. The coil is moving at a speed $v$ in its own plane and away from the wire.

>(a) Find the e.m.f. induced in the coil.

>(b) Let R be the resistance of the coil. Calculate the force needed to move the coil with speed v as described, and show that the mechanical power used to move it is equal to the rate at which heat is generated in the coil.


I have included my workings/thoughts.



I know that i first have to calculate the magnetic field of the wire:
$$B(y)=\mu_0 I/2πy $$
Then the emf,
$$\mathcal{E}=-\frac{dφ(B)}{dt}= -\frac{d}{dt} \int_D^{D+b}B\cdot ds = -\frac{d}{dt}(a\cdot \int_D^{D+b} B\cdot dy)$$


I have been given the answer of
$$\mathcal{E}=\frac{\mu_0Ivab}{2\pi D\left(D+b\right)}$$

What I am having trouble with is the intermediate step getting from the integral to the above answer.

I know that $P_{mech}=F.v$ and that $P_{heat}= v^2/R = \mathcal{E}/R$, but I do not know how to calculate the force on the coil.

 

[TSny]

Hello mrmoriarty and welcome to PF!

$$\int_D^{D+b} B\cdot dy$$

Can you show us your attempt at evaluating this integral?

I know that $P_{mech}=F.v$ ..., but I do not know how to calculate the force on the coil.

There is a basic formula for calculating the magnetic force on a current in a magnetic field.

 

[rude man]

Do you know the Blv law? It's usually better than computing the flux change within the coil. In fact, sometimes where moving media are concerned this approach fails completely.

 

[Zatman]

Apologies for resurrecting the thread, but I have the exact same question set so it seemed pointless to make a new topic.

I have done the first part (calculating the emf), but do not know how to find the force.

I have tried using F = I dl x B with I = emf/R.

F = \frac{\mu_0Iva}{2\pi D}\frac{a}{R}\frac{\mu_0I}{2\pi D} + \frac{\mu_0Iva}{2\pi(D+b)}\frac{a}{R}\frac{\mu_0I}{2\pi(D+b)}

F = (\frac{\mu_0Ia}{2\pi})^2\frac{v}{R}[\frac{1}{D^2}+\frac{1}{(D+b)^2}]

This does not lead to the correct answer. Any help would be appreciated :)

 

[TSny]

Hello, Zatman. Let's break it down.

Let $I'$ be the current induced in the loop. Find an expression for the force on the side of the loop closest to the long wire in terms of $I$ and $I'$ where $I$ is the current in the long wire. Don't bother yet to substitute for $I'$. Just express the force in terms of $I, I', a$ and $D$.

 

[Zatman]

Hi TSny, thanks for your reply. I have:

F = \displaystyle\int^a_0 I'\frac{\mu_0I}{2\pi D}\ dl

F = \frac{\mu_0aII'}{2\pi D}

 

[TSny]

Great. Now do the same for the far side of the loop. Also, think about the directions of the forces. Then combine the two forces.

 

[Zatman]

Okay, so the only things that will be different are
(1) the sign, since the current is in the opposite direction.
(2) D --> D+b

So the force on the far side of the loop is

F_2 = -\frac{\mu_0aII'}{2\pi (D+b)}

The net force is then the sum of the two forces:

F_{net} = \frac{\mu_0aII'}{2\pi}(\frac{1}{D}-\frac{1}{D+b})

F_{net} = \frac{\mu_0aII'b}{2\pi D(D+b)}

 

[TSny]

Fantastic. Now, how would you express the mechanical power required to move the loop at speed v?

 

[Zatman]

P = \frac{dW}{dt} = \frac{d(Fx)}{dt} = F\frac{dx}{dt} = Fv

P = F_{net}v = \frac{\mu_0IvabI'}{2\pi D(D+b)}

Now I think I can substitute the expression for I':

I' = \frac{emf}{R} = \frac{\mu_0Ivab}{2\pi RD(D+b)}

Which gives the correct answer. I see my mistake now; I somehow managed to have two different currents in each section of the wire.

Thanks for your help! :)

 

[TSny]

Good work!

One minor point where you wrote

P = \frac{dW}{dt} = \frac{d(Fx)}{dt} = F\frac{dx}{dt} = Fv.

The middle expression is not actually correct, but the other expressions are correct.

 

[Zatman]

Yes, it is only valid for a constant force? Which in this question is not actually the case, I notice.

 

[TSny]

It would be valid for a constant force, but I don't think it's a good idea to write W = d(Fx)/dt even for that case. During a small time dt, the work done is Fdx. So the rate of doing work is (Fdx)/dt = F(dx/dt) = Fv.