2
$\begingroup$

I am confused by the following example in my textbook.

Question:-

A metallic rod of length $l$ is rotated with a angular velocity $\omega$, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $l$, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field $B$ is present everywhere and is parallel to the axis. What is the emf between the centre and the metallic ring ?

Solution :-

The area of sector traced by the rod in time $t$ is $\dfrac12 l^2 \theta$, where $\theta = \omega t$. Therefore $\varepsilon = \dfrac{B\omega l^2}{2}$.

The math in the solution is not hard to understand. I have trouble understanding the physics.

I know $\varepsilon = \dfrac{\mathrm d\ \phi_B}{\mathrm d \ t} = \dfrac{\mathrm d\ (\bf B\cdot A )}{\mathrm d \ t}$ and in the question it is given that ${\bf B}$ is constant and the angle between $\bf A$ and $\bf B$ is $0$. So we are just left with $\varepsilon = B\dfrac{\mathrm d \ A}{\mathrm d \ t}$.

Why the area taken in the solution is the area of the sector traced by the rod in time $t$ ?

Since the magnetic flux is passing through the whole metallic ring, shouldn't the area be the area of metallic ring ? Which is contant. So the emf should be zero.

$\endgroup$
  • 2
    $\begingroup$ The EMF if proportional to the number of flux lines per second being cut by the moving conductor, which is proportional to the area swept out per second by the moving conductor. $\endgroup$ – John Rennie Oct 15 '17 at 14:09
2
$\begingroup$

To measure the EMF between the axial point and the metallic circumference we have to take a certain point on the circumference. Now, draw a line each from the axial point and the chosen point on the circumference. Join them and connect a galvanometer in parallel. Observe carefully here, w.r.t the point on the circumference, as the rod moves, the area under the circuit changes. So, according to EMI, there is an induced EMF. The change in area is relative. According to you the area was constant because you didn't consider any circuit.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I did not understand any circuit because there not circuit given the example. Doesn't placing galvanometer changes the question asked ? Also if I place galvanometer differently won't I get different answer ? $\endgroup$ – user8277998 Oct 15 '17 at 17:15
  • 1
    $\begingroup$ Well I meant that because you didn't consider any circuit you couldn't visualize the change in area. To measure the EMF between two points you should join them first. You can do that mentally or by drawing a circuit. I said that the change in area was relative. You can take any point on the circumference and just join it with the axial point with a wire. Then you trace the circuit from the centre via the rod to the circumference. From there trace it to the point where the wire is attached and then trace all the way back to the centre via the wire. $\endgroup$ – Amyanshu Jenamani Oct 15 '17 at 17:58
  • 1
    $\begingroup$ But consider this : say the rod starts at 12 o clock and after $t$ seconds goes to 3 o clock. The rod has moved but the area enclosed inside the metallic ring is still $\pi l^2$, so if the magnetic flux is passing through the area $\pi l^2$ in the beginning then it is still passing through the same $\pi l^2$ area after the rod has transversed angular distance $\pi/4$ in clockwise direction. There is no change in area which means the emf should be zero. $\endgroup$ – user8277998 Oct 15 '17 at 18:20
  • 1
    $\begingroup$ The metallic ring is closed curve. How it is not a closed curve ? $\endgroup$ – user8277998 Oct 15 '17 at 18:35
  • 1
    $\begingroup$ You have a open end at the centre, don't you? After all, you have to find induced EMF between the centre and the metallic ring. Just don't think superficialy. Give it some time. $\endgroup$ – Amyanshu Jenamani Oct 15 '17 at 18:37
3
$\begingroup$

There are two possibilities of how you close the loop. (Actually several possibilities, but lets consider two for now ) I have shown them in the picture. In one case, there will be no EMF generated. In other case, there will be EMF generated. Two cases shown in picture below. In both cases, the black shows initial position and blue shows the final position.

enter image description here

On LHS, the voltmeter and wires move with the rod and no EMF is generated. On RHS, the voltmeter does not move and wires stretch. In this case, there will be EMF generated as the area is changing. But in this second case, the math involved is not very easy as we have to find the rate of change of the area of triangle.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Sorry for my poor english !

The above explanations seem very clear to me. It is necessary to close the circuit to make a measurement.

We can, however, give a more formal formulation that takes up the idea of field lines cut by the circuit.

The emf is defined as the circulation of the force per unit of charge, here the magnetic force. When the circuit $CD$ moves during $dt$ the emf is ${{e}_{ind}}=\int\limits_{C}^{D}{(\overrightarrow{v}\wedge \overrightarrow{B})\cdot \overrightarrow{dl}}=\int\limits_{C}^{D}{(\frac{\overrightarrow{\delta r}}{dt}\wedge \overrightarrow{B})\cdot \overrightarrow{dl}}=\frac{1}{dt}\int\limits_{C}^{D}{(\overrightarrow{\delta r}\wedge \overrightarrow{B})\cdot \overrightarrow{dl}}=-\frac{1}{dt}\int\limits_{C}^{D}{\overrightarrow{B}\cdot (\overrightarrow{\delta r}\wedge \overrightarrow{dl})}$

We define the algebraic surface cut by the circuit $\overrightarrow{d{{S}_{c}}}=(\overrightarrow{\delta r}\wedge \overrightarrow{dl})$ and the "cut flux" (in French, "flux coupé"....) $\delta {{\varphi }_{c}}=\int\limits_{C}^{D}{\overrightarrow{B}\cdot (\overrightarrow{\delta r}\wedge \overrightarrow{dl})}$ : the algebraic flux of the magnetic field across the surface crossed by the circuit.

Faraday's law is then ${{e}_{ind}}=-\frac{\delta {{\varphi }_{c}}}{dt}$ and it only leads to Faraday's usual law ${{e}_{ind}}=-\frac{d\Phi }{dt}$ for a closed circuit. In this case, indeed, the magnetic field flux being conservative, $d\Phi =\delta {{\varphi }_{c}}$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

See guys your looking at the problem in the wrong way... The emf can be induced in other ways not just by Faraday's statement about change in the flux.

Imagine carefully now, if the rod is rotating with constant angular velocity in a uniform magnetic field then the particles (infinite) (I would better consider them as free electrons) on the metal rod have different individual velocity vectors. If you apply the the Fleming left hand rule on each of the individual particle you determine the direction of the force as you know the velocity and the magnetic field directions. Now due to this force the free electrons get stagnanted( or increase their concentration) on one of the ends of the rotating rod and die to this equal amount of positive charge on the other end of the rotating rod die to this there will be a potential difference generated between the ends of the rod and if the resistance is not high enough there will even be an induced current between the high and low potentials ( positive charge is high potential and negative charge is low potential)

And for the derivation part can be made simple too I guess, I am sorry I don't have keyboard for which you use to get those symbols but I would love to show you the derivation..

I hope you know that around a rod (the is moving with constant velocity) the emf induced is

$$\mathbf{E} =(\mathbf{v}×\mathbf{B})L$$

Since in this case

$$E=BLv$$

And $v$ is variable and equal to $r\omega$. small emf about small element $dx$

$$dE = B(dx)(\omega x)$$

Integrate and get

$$E= \frac{Bwl^2}{2}$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This question had me puzzled too at some point of time. The way i think about it is the following:

  1. Forget about the circular ring for a minute. You will agree that an emf will be generated across the ends of a rotating rod. But just for the sake of completeness, the electrons in the rod are being forced to move along with the rod. We know that a moving charge experiences a force if placed in a magnetic field(that is not oriented exactly along its direction of motion). In this particular case, the electrons experience a force either towards the center or radially outwards depending upon directions of magnetic field and rotation of rod. That's your emf.

  2. A conductor (not in motion in a magnetic field) will have the same potential everywhere on the surface(and the bulk). So if it was just the circular ring placed in the field with no relative motion whatsoever, the whole of the ring would be at the same potential.

  3. Now we combine the two systems. The potential everywhere on the ring will be the same as the potential being generated on the outer end of the rotating rod. And hence we dont have to even considering worrying about the area of the circle itself. I don't see why the outer ring would make any difference on the outcome or the necessity of thinking about the measuring circuit.

Let me know if you think my explanation is flawed. i will be happy to understand what's actually going on. Thanks.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.