- #1
zenterix
- 488
- 71
- Homework Statement
- Force on an electric charge is given by
$$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})\tag{1}$$
- Relevant Equations
- Emf is defined as the integral of the force per unit charge around a current loop.
$$\mathcal{E}=\oint_C\frac{d\vec{F}}{dq}\cdot d\vec{r}\tag{2}$$
Does this mean we can write the following?
$$\mathcal{E}=\oint_C \vec{E}\cdot d\vec{r}+\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}\tag{3}$$
I haven't seen an equation like the above in my books and notes yet.
What I have seen are two cases.
In one case, we have a uniform magnetic field and we move the current loop in the field in such a way that we have changing magnetic flux.
A simple example of this is a rectangular current loop where one of the sides is movable. This side has velocity ##\vec{v}## and so experiences the effect of a magnetic force ##\vec{v}\times\vec{B}##.
$$\mathcal{E}=\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}\tag{4}$$
Is the ##\vec{E}## term zero in (1) for this case?
In a second case, the conducting loop does not move but the magnetic field is allowed to change in time.
We get an induced emf in the loop but we can't explain it with the Lorentz force.
We explain it with a new law of nature, Faraday's law, which says that
$$\nabla\times \vec{E}=-\frac{\partial\vec{B}}{\partial t}\tag{5}$$
If we integrate both sides along a surface ##S##, then by Stokes' theorem the lhs side is
$$\oint_C \vec{E}\cdot d\vec{s}=\iint_S(\nabla\times \vec{E})\cdot d\vec{a}\tag{6}$$
and so
$$\oint_{C} \vec{E}\cdot d\vec{s}=\iint_S (\nabla\times\vec{E})\cdot\hat{n}da=-\iint_S\frac{\partial \vec{B}}{\partial t}\cdot \hat{n}da\tag{7}$$
Note that the first equality is from Stokes' theorem and the second is from Faraday's law.
We can rewrite as
$$\oint_C \vec{E}\cdot d\vec{s}=-\frac{\partial}{\partial t}\iint_S \vec{B}\cdot \hat{n}da=-\frac{d\Phi_B}{dt}\tag{8}$$
and so
$$\mathcal{E}=-\frac{\partial\Phi_B}{dt}\tag{9}$$
So my question is about comparing the equations
$$\mathcal{E}=\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}$$
$$\mathcal{E}=\oint_C \vec{E}\cdot d\vec{r}$$
Are these just special cases of (3)?
$$\mathcal{E}=\oint_C \vec{E}\cdot d\vec{r}+\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}\tag{3}$$
I haven't seen an equation like the above in my books and notes yet.
What I have seen are two cases.
In one case, we have a uniform magnetic field and we move the current loop in the field in such a way that we have changing magnetic flux.
A simple example of this is a rectangular current loop where one of the sides is movable. This side has velocity ##\vec{v}## and so experiences the effect of a magnetic force ##\vec{v}\times\vec{B}##.
$$\mathcal{E}=\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}\tag{4}$$
Is the ##\vec{E}## term zero in (1) for this case?
In a second case, the conducting loop does not move but the magnetic field is allowed to change in time.
We get an induced emf in the loop but we can't explain it with the Lorentz force.
We explain it with a new law of nature, Faraday's law, which says that
$$\nabla\times \vec{E}=-\frac{\partial\vec{B}}{\partial t}\tag{5}$$
If we integrate both sides along a surface ##S##, then by Stokes' theorem the lhs side is
$$\oint_C \vec{E}\cdot d\vec{s}=\iint_S(\nabla\times \vec{E})\cdot d\vec{a}\tag{6}$$
and so
$$\oint_{C} \vec{E}\cdot d\vec{s}=\iint_S (\nabla\times\vec{E})\cdot\hat{n}da=-\iint_S\frac{\partial \vec{B}}{\partial t}\cdot \hat{n}da\tag{7}$$
Note that the first equality is from Stokes' theorem and the second is from Faraday's law.
We can rewrite as
$$\oint_C \vec{E}\cdot d\vec{s}=-\frac{\partial}{\partial t}\iint_S \vec{B}\cdot \hat{n}da=-\frac{d\Phi_B}{dt}\tag{8}$$
and so
$$\mathcal{E}=-\frac{\partial\Phi_B}{dt}\tag{9}$$
So my question is about comparing the equations
$$\mathcal{E}=\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}$$
$$\mathcal{E}=\oint_C \vec{E}\cdot d\vec{r}$$
Are these just special cases of (3)?