How do I find the Direction of an induced electric field?

[Ahmed1029]

Faraday's law tell's you about the line intergal of the electric field, but you have to know the direction of the induced electric field first in order to properly apply it. How can I find its direction? Is it in the same direction as the induced current?

 

[BvU]

Wouldn't that be Lenz's law ?

 

[Ahmed1029]

Wouldn't that be Lenz's law ?

View attachment 313360

Lenz's law determines the direction of the current. Is the induced field exactly in the direction of the current?

 

[topsquark]

Lenz's law determines the direction of the current. Is the induced field exactly in the direction of the current?

Hint: F = qE. The current due to an electric field is the flow of charges. So the field would have to either be in the direction of the flow or in opposite direction. What is the charge of your charged particles?

-Dan

 

[hutchphd]

Or perhaps more directly $$\mathbf J =\sigma \mathbf E $$ using current density and conductivity

 

[Ahmed1029]

I know within the current-occupied volume it has to be this way, but what about other points?

 

[topsquark]

I know within the current-occupied volume it has to be this way, but what about other points?

What is the direction of the field at other points? Are you asking about what happens outside the region of flux change?

-Dan

 

[Ahmed1029]

What is the direction of the field at other points? Are you asking about what happens outside the region of flux change?

-Dan

No, I meant there is a changing magnetic field outside the interior of the cable itself which would induce a curly electric field. But I realized it was unnecessary to know this piece of detail; Your previous answers suffice.

 

[vanhees71]

The integral forms of Maxwell's equations are always to be used with some care to get the signs right. In that sense they are more complicated than the fundamental differential forms of the law. So we start from the simple differential form of Faraday's law of induction (in SI units)
$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$
You get the integral law by using Stokes's integral theorem, i.e., you integrate the equation over an oriented surface, $A$, i.e., a surface with surface-normal vectors $\mathrm{d}^2 \vec{f}$ oriented arbitrarily in one of two possible directions. Then on the left-hand side you can transform the surface integral as a line integral along the boundary, $\partial A$, of the surface. Its orientation must be chosen using the right-hand-rule relative to the orientation of the surface-normal vectors. Then you get
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Finally, if we assume for simplicity that the surface and its boundary are at rest (and ONLY then) you can take the time derivative out of the surface integral
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

 

[Ahmed1029]

The integral forms of Maxwell's equations are always to be used with some care to get the signs right. In that sense they are more complicated than the fundamental differential forms of the law. So we start from the simple differential form of Faraday's law of induction (in SI units)
$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$
You get the integral law by using Stokes's integral theorem, i.e., you integrate the equation over an oriented surface, $A$, i.e., a surface with surface-normal vectors $\mathrm{d}^2 \vec{f}$ oriented arbitrarily in one of two possible directions. Then on the left-hand side you can transform the surface integral as a line integral along the boundary, $\partial A$, of the surface. Its orientation must be chosen using the right-hand-rule relative to the orientation of the surface-normal vectors. Then you get
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Finally, if we assume for simplicity that the surface and its boundary are at rest (and ONLY then) you can take the time derivative out of the surface integral
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

But you will have to know the direction of the electric field beforehand in order to apply that last equation in a way that gets you the electric field, which is what I'm asking.

 

[vanhees71]

In general you cannot get $\vec{E}$ and $\vec{B}$ so easily from the integral laws. You only can get such solutions under the assumption of very symmetric situations (e.g., the electrostatic Coulomb field from rotational symmetry around the resting point charge as a source).