Electromagnetic induction (Faraday's law)

[Meow12]

Homework Statement

A wire is bent to contain a semi-circular curve of radius 0.25m. It is rotated at 120rev/min as shown into a uniform magnetic field below the wire of 1.30T. What is the maximum emf induced between the left and right sides of the wire in V?

Relevant Equations

$\phi=\omega t$
$\Phi_B=BA\cos\phi$
$\displaystyle\epsilon=\oint\vec E\cdot d\vec{l}=-\frac{d\Phi_B}{dt}$

I used the formulas above and got the right answer:

$\phi=4\pi t$
$\Phi_B=0.128\cos 4\pi t$
$\epsilon=1.61\sin 4\pi t$
$\epsilon_\text{max}=1.61\ V$

But I still have a question: Faraday's law requires a closed loop, whereas the semi-circular loop I used was open on one side.

 

[BvU]

Underlying phenomenon for Faraday's law is the Lorentz force. The steps from there to Faraday's law and to your exercise answer are the same.

$\$

 

[kuruman]

Where did you get the expression that you used? Under what conditions is it applicable? Just because it gives the "right answer" does not justify its use. For example, when the semicircle is in the orientation shown in the figure, the flux through the semicircle is zero. furthermore, it is zero not instantaneously as your expression predicts, but half the time needed for a complete revolution while the semicircle is outside the field.

If you want to apply Faraday's law, do it right. Consider a closed loop as shown below and use the definition of magnetic flux, $$\displaystyle\epsilon=\oint\vec E\cdot d\vec{l}
=-\frac{d}{dt}\left(\mathbf{B}\cdot \mathbf{\hat n}~da\right).$$ Since the magnetic field is constant and uniform, the only time dependence is in the normal to the loop $\mathbf{\hat n}$ so you have to find an expression for $\dfrac{d \mathbf{\hat n}}{dt}$ and take the dot product.

Alternatively, you can consider this to be a motional emf problem and find the potential difference between the two ends of the semicircle.

 

[Delta2]

I am not quite sure about @kuruman suggestion of the closed loop. I would choose as closed loop the semicircle together with the diameter. Then as @kuruman suggest i would view it as a motional EMF problem and compute the motional EMF as the line integral $$EMF=\int_C (\vec{B}\times\vec{v})\cdot \vec{dl}$$ where the curve of integration $C$ is the aforementioned closed loop. It is easy to see that $\vec{B}\times\vec{v}=0$ for the points of the closed loop that lie on the diameter because $v=0$ there.

You might tell me that there is no conducting path in the diameter but even if there was, the EMF would be the same for the whole closed loop, the only difference in physics is that we would have current circulating in that hypothetical closed loop.

 

[kuruman]

I am not quite sure about @kuruman suggestion of the closed loop. I would choose as closed loop the semicircle together with the diameter.

We are saying the same thing. If you consider the formal integral over the entire circle, only the semicircle that has field lines going through it contributes non-zero terms while the other half contributes a bunch of zeroes. So one gets the same answer as using the semicircle that you suggest.

 

[Delta2]

We are saying the same thing. If you consider the formal integral over the entire circle, only the semicircle that has field lines going through it contributes non-zero terms while the other half contributes a bunch of zeroes. So one gets the same answer as using the semicircle that you suggest.

Ehm this isnt so clear to me, with your choice of closed loop, the way i see it, we have EMF during the whole rotation time (full cycle) (during half cycle the one half semicircle contributes non zero, during the next half cycle the other half semicircle contributes non zero), while with my choice we have EMF only during half cycle.

 

[kuruman]

You are right. I don't know what I was thinking.

 

[kuruman]

Actually, there is no reason to consider a closed loop enclosing an area for a formal application of Faraday's law or using motional emf. We are looking for the maximum emf. Since $\text{emf}=-\dfrac{d\Phi_M}{dt},$ all one has to do is find an expression for $\Phi_M(t)$ and maximize its time derivative.