Find Induced Emf in this Coil

[ltkach]

A flat, rectangular coil of dimensions L and w is pulled with uniform speed v through a uniform magnetic field B with the plane of its area perpendicular to the field (the figure (Figure 1) ).How is the emf zero?

My reasons as to why it's zero:

the trig part forces the problem to be zero?

Emf Induced ===== (Enclosed Integral) (vBsin(90))*dl*cos(90) = 0

Thanks!

 

[ltkach]

I think I figured it out.

I have attached a photo of what I did.

 

[disknoir]

Think about the Lorentz force on the charge carriers. The direction of v x B in the opposite sides of the loop will oppose each other so there is no current induced.

Hope I understood the question correctly; I'm using the iPhone app, and I'm unable to view your pictures on my phone. :)

 

[ltkach]

can you check out the third picture i posted. I four different integrals. they all canceled out to be zero.

 

[Mr.Atomizer]

A non zero EMF can only occur if there is a change in flux. Since the magnetic field here is constant, the area of the coil doesn't change and the orientation of the coil with respect to the magnetic field doesn't change, there is no change in flux and hence no induced EMF produced.

 

[ltkach]

Mr.Atomizer,

What about the equation for non stationary magnetism.
Emf= integral of vXB(dot)dL

 

[ltkach]

Non stationary Induced Emf sorry.

 

[vanhees71]

The electromotive force is 0 here. So your calculation is correct. This can also be immediately seen from Faraday's Law in integral form. To derive it one must be a bit carefull. It's wrong in many textbooks.

One always should start from the local (differential) form of the Lorentz invariant Maxwell equations. The Faraday Law in any inertial frame of reference reads
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
Now let $A$ be any area (time dependent or not) and $\partial A$ its boundary. Further let the relative orientation of the surface elements and the boundary curve be defined in the standard way by the right-hand rule. Then, according to Stokes's theorem you have
$$\int_{\partial A} \mathrm{d} \vec{x} \cdot \vec{E}=-\int_A \mathrm{d} \vec{A} \partial_t \vec{B}. \qquad (*)$$
Now the point is to bring the partial time derivative out of the integral. This, however gives two contributions: Besides the partial time derivative of the time-dependent magnetic field, it also contains a piece from the change of the moving surface. The result of the analysis is
$$\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d} \vec{A} \cdot \vec{B}=\int_A \mathrm{d} \vec{A} \cdot \partial_t \vec{B}-\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}).$$
Here $\vec{v}=\vec{v}(t,\vec{r})$ is the velocity of the boundary-line element at the time and position $(t,\vec{r})$. The proof is given in the Wikipedia:

http://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof_of_Faraday.27s_law

You have to open the box to see the proof.

Now we have the magnetic flux defined by the integral on the left-hand side,
$$\Phi(t)=\int_{A} \mathrm{d} \vec{A} \cdot \vec{B},$$
and then Faraday's Law (*) tells us
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=-\frac{\mathrm{d} \Phi}{\mathrm{d} t}.$$

In your case, because the magnetic field is homogeneous, the magnetic flux through the coil doesn't change with time and thus the EMF, i.e., the line integral on the left-hand side of the above equation, vanishes.