We explain that motional emf is induced because charges in a conductor move along with the conductor and as a result a magnetic force pushes them to one side of the conductor. But in case of a stationary conductor with v=0 how can variable magnetic field apply force on stationary charges inside a stationary conductor? If we say charges (electrons in this case) are in random motion so they undergo magnetic force, in such case , constant magnetic field should also induce emf? Thanks if u answer my question.
[Physics] Induced emf in stationary conductor lying in varying magnetic field
electromagnetism
Related Solutions
An EMF from a source is defined as a force per unit charge line integrated about the instantaneous position of a thin wire so for an electromagnetic source:
$$\mathscr E=\oint_{\partial S(t_0)} \left(\vec E + \vec v \times \vec B\right)\cdot d \vec l.$$
Where $S(t_0)$ is a surface enclosed by the wire at time $t=t_0$ and the partial means the boundary, so $\partial S(t_0)$ is the instantaneous path of the wire itself at $t=t_0.$ The $\vec v$ is the velocity of the actual charges. Note this is not necessarily the work done on the charges if the wire is moving since the wire goes in a different direction than the charges go when there is a current.
Now, if the wire is thin and the charge stays in the wire and there are no magnetic charges we get $$-\oint_{\partial S(t_0)} \left(\vec v \times \vec B\right)\cdot d \vec l=\frac{d}{dt}\left.\iint_{\partial S(t)}\vec B(t_0)\cdot \vec n(t)dS(t)\right|_{t=t_0}$$
And regardless of magnetic charges or thin wires or whether charges stay in the wires we always get $$\oint_{\partial S(t_0)} \vec E\cdot d \vec l=\iint_{S(t_0)}\left.-\frac{\partial \vec B(t)}{\partial t}\right|_{t=t_0}\cdot \vec n(t_0)dS(t_0).$$
So combined together we get:
$$\mathscr E=\oint_{\partial S(t_0)} \left(\vec E + \vec v \times \vec B\right)\cdot d \vec l=-\left.\left(\frac{d}{dt}\Phi_B\right)\right|_{t=t_0}$$
The force due to the motion of the wire is purely magnetic, and the force due to the time rate of change of the magnetic field is purely electric. And the work done is an entirely different question than the EMF. The work happens for a motional EMF when a Hall voltage is produced.
So,is the former case of when the loop moves in a stationary magnetic field different?
A moving wire feels a magnetic force and magnetic forces can be a source term in an EMF.
Is electric field in the loop due to "motional emf" conservative?
Motional EMF is not caused by electric forces, it is caused by magnetic forces. Since magnetic forces depend on velocity, the word conservative does not even apply since the force depends on the velocity, not merely the path, and they don't do work.
And the book also,at one point, expresses electric field due to motional emf as a scalar potetnial gradient.
If the wire develops a Hall voltage due to the magnetic force, then the charge distribution for the Hall voltage would set up an electrostatic force, which is conservative.
In particular, if the magnetic field is not changing, then the electric field is conservative.
However,motional emf does sounds similar to induced emf.
When you compute the magnetic flux at two times the term $-\vec B \cdot \hat n dA$ can change for two reasons, a changing loop and a time changing magnetic field. You really get both effects from the product rule for derivatives. The one from the time changing magnetic field becomes equal to the circulation of the electric force per unit charge. The one from the time changing loop becomes equal to the circulation of the magnetic force per unit charge.
My question is,is E due to motional emf and induced E different or not,and why so?
The electric field is conservative if the magnetic field is not changing in time. And if the magnetic field is not changing in time, the EMF is due solely to the moving charges in the moving wire interacting with a magnetic field.
- The case of a moving conductor and a stationary conductor is fundamentally different. When the conductor is stationary, a changing magnetic field produces an electric field everywhere in space, whose curl along any loop enclosing the varying magnetic field is non-zero, given by $\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}$. Using Stokes law, we easily find the emf to be the rate of change of flux through the loop. NOTE: it is the electric field produced in space that is the cause of emf(in this case) not any force.
The emf across an open ended conductor(suppose the ends are A and B) is ONLY due to the conservative electrostatic field produced by the separated charges, due to the magnetic force, given by the integral of the electrostatic field over the length of the conductor for A to B : $\mathcal E=-\int \mathbf E_s\cdot d\mathbf l$. As the conductor speeds up, due the greater magnetic force, greater charge is deposited at the ends, resulting in greater magnitude of the electrostatic field ($\mathbf E_s$) across the conductor, which in turn increases its integral over the length (which is the emf of course!).
Again, for an open-ended conductor, we take the electrostatic field, which is $qvB/q = vB$. Substituting In the formula and integrating, $\mathcal E = -vBL$. So, the formula remains valid.
You are again confusing the non-conservative electric field $\mathbf E$ with the conservative electrostatic field ($\mathbf E_s$). When the circuit is closed, the emf: $\mathcal E=-\int \mathbf E\cdot d\mathbf l$, where $\mathbf E$ is the total field( due to electric as well as electrostatic over the whole loop). The electrostatic part is obviously zero over the whole loop (because it is conservative), but the electric field is given by Maxwell’s Equation: $\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}$. For constant $\mathbf B$, $\nabla\times\mathbf E=0$, which from Stokes law suggests that its integral around the loop is also zero, producing zero emf. For varying magnetic field, the integral of electrostatic field vanishes anyway, but $\nabla\times\mathbf E$ is not zero, which renders the emf non zero in the integral. Do not bring in electrostatic field in emf in a closed circuit. You’ll confuse yourself.
Best Answer
For the stationary case with time varying magnetic field Maxwell's equations produce an induced electrical field which will accelerate charges. For the case of moving conductor and stationary field, a Lorentz transform of the constant magnetic field into the frame of the moving conductor will, again, produce an induced electrical field. In both cases the forces on charges are caused by the electric field component.
A constant magnetic field in the coordinate system of moving charges will therefor always produce an effective electric field component, which will result in an acceleration of charged particles. This can be shown very nicely in experiments, in which electron beams get deflected by a constant magnetic field.