According to Faraday, a change in magnetic flux induces current in a conductor. Just suppose we take a circular coil, and pass a magnetic field inside it, and keeping changing the direction of magnetic field from 0 and 180 degrees, without change it's magnitude ( Suppose the change in direction is instantaneous ). Since flux is the number of lines passing perpendicularly through a given area, which is obviously not changing, is it safe to say that no current will be induced in the coil while doing this?
1
-
$\begingroup$ You have two opposite questions in the title and the body. $\endgroup$– JasperCommented Dec 1, 2018 at 13:13
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
0
(As pointed out in a comment below, the question is asked differently in the title and in the body. My answer is written taking into account the body, therefore the answer to the body is no, while the answer to the original title is yes.)
Short answer: no, it is not safe, since you are not considering the orientation of these lines with respect to the surface.
Long and more detailed answer:
When calculating any flux (in this case, for the sake of simplicity, I will calculate it for the vector field $\mathbf{B}$, to be consistent with the notation of the magnetic field), from a mathematical point of view, you are computing the following integral: $$ \Phi = \int_{\Sigma} \mathbf{B} \cdot d\boldsymbol\sigma $$ where I considered $\Sigma$ to be the surface across which you are calculating the flux and $d\boldsymbol\sigma$ the oriented infinitesimal surface. Defining a normal $\mathbf{n}_\sigma$ to that surface, we can rewrite this integral as: $$\Phi = \int_{\Sigma} \mathbf{B} \cdot \mathbf{n}_{\sigma}\, d\sigma $$ and, therefore, depending on the orientation of the vector field $\mathbf{B}$ (assuming the surface $\Sigma$ across which you are evaluating the flux to not be changing) you will obtain a different result of the scalar product and, consequently, a different flux.
In your specific case, I can assume for example to be evaluating the flux across a surface whose normal is parallel (and with the same direction) of the initial magnetic field $\mathbf{B}_0$. It will thus give rise to a certain flux $\Phi_0$. When the orientation of the magnetic field suddenly changes: $$ \mathbf{B}_1 = -\mathbf{B}_0,\quad |\mathbf{B}_1| = |\mathbf{B}_0| $$ this will change the sign of the scalar product inside the integral and, consequently, this will change the sign of the flux: $$ \Phi_1 = -\Phi_0. $$ If, as you assumed, this change is almost instantaneous (thus being non-physical), we can describe it with a step distribution and it will thus give rise to a step in the representation of the flux. Assuming $t_0$ to be the time instant associated to this change: $$ \Phi(t) = \begin{cases} \Phi_0,\quad& t<t_0 \\ \Phi_1,\quad & t>t_0\end{cases}. $$ Since the induced current is proportional, from the Faraday-Neumann-Lenz law, to the time derivative of this flux, then we will have a current that will be proportional to a Dirac delta distribution: $$ I(t) \propto \delta(t-t_0) $$ for each flip of the magnetic field, thus for sure not being negligible.
The key concept, in this case, is that the flux is computed across an oriented surface.
As a side note on the importance of these currents, consider a material with a finite electric resistivity: in this case, every single flip of the magnetic field will imply a dissipation of energy in the material due to the Joule effect. This means that each flip of the magnetic field will actually "cost some energy" to the system, therefore it will be more or less hard to flip the sign of the magnetic field. Moreover, it is impossible to have an instantaneous flip since it will imply a current that is pointwisely diverging in the time instant $t_0$, thus dissipating into heat an infinite amount of energy in that single time-instant. In general, every change of this magnetic field will be "smooth and gentle" (not really mathematical as a description, but quite intuitive), therefore also the induced current will last longer than a single time instant.
$\begingroup$
$\endgroup$
Can Induced current be obtained by only changing the direction of the Magnetic Field?
The answer is yes. If we temporarily ignore the condition that the change in the magnetic field be instantaneous, we can generate an oscillating current by spinning a permanent magnet near a coil in such a way that the magnet's North and South poles are changing places as the magnet rotates.
Since flux is the number of lines passing perpendicularly through a given area...
The "area" here is an oriented two-dimensional region of space with a specific outline. When the magnet is rotated about a generic axis, the magnetic flux through that region does change (changing smoothly from positive to negative and back), and an electric field is induced as a result. If the outline is occupied by a physical loop of conducing wire, then that electric field will induce a current.
In the mathematical limit as this physically-realizable smoothly changing field approaches an instantaneous change, the current approaches zero at all times except right when the change is occurring. This is explained nicely in Jackl's answer.
answered Dec 1, 2018 at 12:45
$\begingroup$
$\endgroup$
The time derivative in your example is undefined and so is the EMF on the loop. The answer is:no it is not safe to say that the EMF will be zero. In fact any value it possible except zero.
$\begingroup$
$\endgroup$
Great Java web app to illustrate this: Faraday's Electromagnetic Lab.
Five programs illustrating Faraday's Law.
Select Pickup Tab. Move coil or Magnet. Place Magnet in coil. Click on Flip Polarity. Repeat Flip Polarity.
As you do this a voltage will be induced in coil. Lightbulb will flash or Meter will deflect. This only happens for instance magnet is flipped.
If you move magnet or coil and you will get induced voltage.
Faraday's Law states: Whenever the flux linked or associated with a circuit changes, a voltage is induced in the circuit.
It's not easily flipped, but if you could do it. Flux $\phi$ has changed so a voltage must be induced.
$$V_{Induced} = N B \ell \mathscr{v}$$
where: N = Number of turns on coil.
B = Flux density in T.
$\ell$ = Length of coil in m.
$\mathscr{v}$ = Velocity in m/s.
Faraday's Electromagnetic Lab allows all of these parameters to be adjusted. All are directly proportional to induced voltage,
Flipping the magnet instantly is impossible, but rotating it is. The last Tab is a Generator, which will illustrate this.
But you can flip DC current in an electromagnet, which would be your concept.
