Problem on induced EMF

[arpon]

Homework Statement

[/B]
Suppose, there is a 'constant' magnetic field $B$ in the upward direction. A loop of conductive wire(XYZ in the picture, which is connected to a closed circuit with resistance R) is placed horizontally on it. The area of the loop is being increased with time. Will there be any current flow in the circuit?

Homework Equations

$\mathcal {E} = -\frac{d \phi}{dt}$
$\phi = BA cos \theta$

The Attempt at a Solution

I think, as the area is changing, there will be a current flow proportional to the rate of change of area inclosed by the loop.

 

[BvU]

It's good to think of this induced emf as a consequence of the Lorentz force on the charge carriers in the wire. For the loop the velocity of the wire segments is outwards, perpendicular to B. The Lorentz force is $\propto \vec v \times \vec B$, so clockwise.

The xyz directions require some explanation...
And $\Delta \Phi$ due to moving the outer loop into the B field (and changing its area as well ?) is a complication I disregarded.

 

[arpon]

And ΔΦ \Delta \Phi due to moving the outer loop into the B field (and changing its area as well ?) is a complication I disregarded.

Can't I calculate the induced EMF in this way? :
$\mathcal E = \frac{d\Phi}{dt} = \frac{d(BA)}{dt} = B \frac{dA}{dt}$

The Lorentz force is ∝v⃗ ×B⃗ \propto \vec v \times \vec B, so clockwise.

And how can I calculate the induced EMF or current flow by the formula you mentioned: $\vec F = q \vec v \times \vec B$ [$q$ is the total charge of the charge carriers in the wire]

The xyz directions require some explanation...

What kind of explanation?

 

[BvU]

Can't I calculate the induced EMF in this way? :
$\mathcal E = \frac{d\Phi}{dt} = \frac{d(BA)}{dt} = B \frac{dA}{dt}$

Yes you can. dA is for the entire loop, not just the circle part.

And how can I calculate the induced EMF or current flow by the formula you mentioned: $\vec F = q \vec v \times \vec B$ [$q$ is the total charge of the charge carriers in the wire]

Hard work. It's done for you here . Wire is neutral, of course. But there are mobile charge carriers and immobile ones. You refer to the former, of course.

What kind of explanation?

A little better than just writing x, y and z around the circular loop. You do indicate that B is in the z direction. Add that the circular loop is in the xy plane. (And you can avoid the complication I mentioned by placing the rectangular part of the loop in the yz plane, so that that part of $\Phi=0$

 

[arpon]

A little better than just writing x, y and z around the circular loop. You do indicate that B is in the z direction. Add that the circular loop is in the xy plane. (And you can avoid the complication I mentioned by placing the rectangular part of the loop in the yz plane, so that that part of Φ=0

Actuallly, I named the points on the circular loop as x,y,z ; I realize it was not wise, because it creates confusion with the x,y,z-axes. However, I am explaining it again, the magnetic field is in the Z direction. The rectangular part of the circuit is on the yz-plane, and the circular part(which is changing the area enclosed by itself) is on the xy-plane, i.e., perpendicular to the direction of magnetic field.
Now, I would request you to revise your posts keeping in mind this explanation.

 

[BvU]

As far as I can distinguish the way you intended the configuration is indeed the way I understood it. So: yes, the movement of the wire increasing the diameter of the loop induces an electromotive force. And it can be calculated using Faraday's law $$
\mathcal E = \oint_{\partial \Sigma} \vec E \cdot d\vec \ell= -{d\over dt}\;\int_\Sigma \vec B \cdot \vec{dA}
$$which comes down to $B\; \frac{dA}{dt}$