Will the induced electric field be circumferential?

[Adesh]

Homework Statement

Let’s say we have a right handed Cartesian system and magnetic field goes in positive z direction.

Relevant Equations

$\oint \vec E \cdot d\vec l = \frac {d\phi_B}{dT}$

Let’s say we have a right handed Cartesian system and magnetic field goes in positive z direction, and let’s assume that the magnitude of magnetic field varies with time.

Now, if I draw a circle with radius $r$ in the $x-y$ plane and let the magnetic field pass through it and vary with time, then by laws of electrodyanmics we have $$\oint \vec E \cdot d\vec l = \frac {d\phi_B}{dt}$$
My question is can I write $\oint \vec E \cdot d\vec l = E ~ 2\pi r$ ? That is can I assume induced electric field to be circumferential?

 

[rude man]

Homework Statement:: Let’s say we have a right handed Cartesian system and magnetic field goes in positive z direction.
Relevant Equations:: $\oint \vec E \cdot d\vec l = \frac {d\phi_B}{dT}$

Let’s say we have a right handed Cartesian system and magnetic field goes in positive z direction, and let’s assume that the magnitude of magnetic field varies with time.

Now, if I draw a circle with radius $r$ in the $x-y$ plane and let the magnetic field pass through it and vary with time, then by laws of electrodyanmics we have $$\oint \vec E \cdot d\vec l = \frac {d\phi_B}{dt}$$
My question is can I write $\oint \vec E \cdot d\vec l = E ~ 2\pi r$ ? That is can I assume induced electric field to be circumferential?

Yes, if your B field is also cicular.
BTW you should have a - sign in front of $d\phi/dt$.

 

[Adesh]

M

Yes, if your B field is also cicular.
BTW you should have a - sign in front of $d\phi/dt$.

magnetic field is going up in the z direction.

 

[rude man]

magnetic field is going up in the z direction.

Right. So the current is going clockwise. That makes it negative.
Why: It's based on Stokes's theorem which says that the net curl of a vector field normal (pointing outward) to a closed surfac S integrated over that surface equals the counterclockwise circulation of that vector around any aperture also with outward pointing normal. The right-hand rule applies to the circulation sign.

In other words, $\iint_S \nabla x \mathbf E \cdot dS = \oint \mathbf E \cdot dl$ with the right-hand rule. So a positive current would have to be counterclockwise but yours is clockwise so tha makes it negative. Remember, the relevant maxwell equation is $\nabla x \mathbf E = -\partial B/\partial t$.

 

[Adesh]

Right. So the current is going clockwise. That makes it negative.
Why: It's based on Stokes's theorem which says that the net curl of a vector field normal (pointing outward) to a closed surfac S integrated over that surface equals the counterclockwise circulation of that vector around any aperture also with outward pointing normal. The right-hand rule applies to the circulation sign.

In other words, $\iint_S \nabla x \mathbf E \cdot dS = \oint \mathbf E \cdot dl$ with the right-hand rule. So a positive current would have to be counterclockwise but yours is clockwise so tha makes it negative. Remember, the relevant maxwell equation is $\nabla x \mathbf E = -\partial B/\partial t$.

From $$\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}$$ we can infer that the curl points in the negative z direction therefore electric field rotate in the x-y plane. My question is around which point does the vector field $E$ going to swirl? Is it with respect to origin?

 

[vishwas]

swirling takes place around area associated with flux change taking center of that area as origin
yet electric field and changing MF are in independent planes

 

[Adesh]

yet electric field and changing MF are in independent planes

Didn't get you? Can you please explain a little more?

 

[vishwas]

well electric and magnetic fields can be in independent planes in 3 dimension
but will feel each others effect
i added this because it may confuse you that when magnetic field fluctuate then electric field swirls and can somehow intersect the magnetic plane
thats not possible right??

 

[Adesh]

well electric and magnetic fields can be in independent planes in 3 dimension
but will feel each others effect
i added this because it may confuse you that when magnetic field fluctuate then electric field swirls and can somehow intersect the magnetic plane
thats not possible right??

Yes, but how can I prove that the induced electric field going to go round and round about the origin. I mean can i get an equation for electric field just by the informations that I have given in my first post.

 

[vishwas]

this might explain why electric field curves

 

[Adesh]

It is given that the direction of magnetic field is in positive $z$ direction and with time it's magnitude is changing.

Given only the above information we can write $$ \nabla \times \vec E = \frac{- \partial \vec B}{\partial t}$$ now if write the above equation component wise and keeping in mind that the x and y component of magnetic field is zero, we have $$ \left( \nabla \times \vec E \right) _x = 0 \\
\left ( \nabla \times \vec E \right)_y = 0 \\
\left ( \nabla \times \vec E \right)_z = \frac{-\partial B}{\partial t}$$ now writing out the the components of curl and keeping in mind that the $z$ component of $\vec E$ will be zero because curl of $\vec E$ points in negative z direction and hence $\vec E$ needs to be perpendicular to it's curl. $$ \left ( \nabla \times \vec E \right)_x = 0 \implies \frac{\partial E_y}{\partial z} = 0 \\
\left ( \nabla \times \vec E \right) = 0 \implies \frac{\partial E_x}{\partial z} = 0 \\
\left ( \nabla \times \vec E \right)_z = -\frac{\partial B}{\partial t} \implies \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} = \frac{-\partial B}{\partial t}$$

Now, how can we proceed after this? Is it possible to extract any more information about the electric field?

 

[rude man]

It is given that the direction of magnetic field is in positive $z$ direction and with time it's magnitude is changing.

Given only the above information we can write $$ \nabla \times \vec E = \frac{- \partial \vec B}{\partial t}$$ now if write the above equation component wise and keeping in mind that the x and y component of magnetic field is zero, we have $$ \left( \nabla \times \vec E \right) _x = 0 \\
\left ( \nabla \times \vec E \right)_y = 0 \\
\left ( \nabla \times \vec E \right)_z = \frac{-\partial B}{\partial t}$$ now writing out the the components of curl and keeping in mind that the $z$ component of $\vec E$ will be zero because curl of $\vec E$ points in negative z direction and hence $\vec E$ needs to be perpendicular to it's curl. $$ \left ( \nabla \times \vec E \right)_x = 0 \implies \frac{\partial E_y}{\partial z} = 0 \\
\left ( \nabla \times \vec E \right) = 0 \implies \frac{\partial E_x}{\partial z} = 0 \\
\left ( \nabla \times \vec E \right)_z = -\frac{\partial B}{\partial t} \implies \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} = \frac{-\partial B}{\partial t}$$

Now, how can we proceed after this? Is it possible to extract any more information about the electric field?

We "proceed" by invoking the Stokes theorem which relates the surface integral of $\nabla x \mathbf E$ wth the closed loops of $\mathbf E$ which are given by $\oint \mathbf E \cdot d\mathbf l = \iint_S \nabla x \mathbf E \cdot d \mathbf S = -\frac {\partial \phi} {\partial t}$.

The E field is circular only if the B field is also circular. If you center the B field at x=y=0 then the E field is circular around the origin. If the B field is irregular then all you know is that the circulation of $\mathbf E = -d\phi/dt$. The E field still comes in closed loops but they will then be of irregular shape But $\nabla x \mathbf E = - dB/dt$ is a point expression and holds for all points, whether they be inside or outside a B field.

 

[Adesh]

The E field is circular only if the B field is also circular.

I'm having trouble in understanding the above part of your reply. B field is going straight downwards so how can it be circular? I request you to please clear my doubt.

If you center the B field at x=y=0 then the E field is circular around the origin.

How can be prove this (I mean the above quote) ?

 

[rude man]

I'm having trouble in understanding the above part of your reply. B field is going straight downwards so how can it be circular? I request you to please clear my doubt.

Think of the B field as a right circular cylindrical shape with principal axis the z axis and centered at the origin x=y=0.

How can be prove this (I mean the above quote) ?

The B field is circular in the x-y plane centered at x=y=0. The circulation is $\oint \mathbf E \cdot d\mathbf l = -dB/dt$. Then by symmetry the contours must be circles.

What else could they be?

 

[Adesh]

Sir I’m sorry for asking things again and again. What is meant by contour?

 

[rude man]

Sir I’m sorry for asking things again and again. What is meant by contour?

A closed line.
For example, a circle is a contour. An ellipse is a contour. A hyperbola is not a contour, it does not close on itself.
A contour can have any shape, in 2 or 3 dimensions.
Do you have an English dictionary?

 

[hutchphd]

The E field is circular only if the B field is also circular.

This is at best badly stated. Are you truly describing a circular B field? I think you mean to describe a cylindrically symmetric field pointing along the z-axis.
Also the term "contour" means a constant level of a scalar field to me. The term contour is seldom used to describe a vector field directly (in my physics vernacular).

 

[Adesh]

This is at best badly stated. Are you truly describing a circular B field? I think you mean to describe a cylindrically symmetric field pointing along the z-axis.
Also the term "contour" means a constant level of a scalar field to me. The term contour is seldom used to describe a vector field directly (in my physics vernacular).

How cylindrical B field ensures that E field is going to go in circles around the origin? I mean when we write $$\oint \mathbf E \cdot d\mathbf l = -\frac{d\phi_B}{dt}$$ How it tells us that the Electric field is swirling in circles?

 

[rude man]

How cylindrical B field ensures that E field is going to go in circles around the origin? I mean when we write $$\oint \mathbf E \cdot d\mathbf l = -\frac{d\phi_B}{dt}$$ How it tells us that the Electric field is swirling in circles?

I will try one last time:
You know the E field comprises closed lines i.e. closed contours.
If the B field is circular what ELSE can these contours be besides circles?
Look at the picture and try to draw a closed contour other than a circle, then try to justify for yourself how that contour could possibly be generated in view of the symmetry of the B field.

 

[Adesh]

I will try one last time:
You know the E field comprises closed lines i.e. closed contours.
If the B field is circular what ELSE can these contours be besides circles?
Look at the picture and try to draw a closed contour other than a circle, then try to justify for yourself how that contour could possibly be generated in view of the symmetry of the B field.

All right! Thanks.

 

[hutchphd]

If the B field is circular what ELSE can these contours be besides circles?

What does this mean??A circular B field is produced by , for instance, a long straight wire. I don't know what you are talking about here, nor does the OP.

 

[hutchphd]

Again I believe part of the issue here is semantic...circular B field is ill-defined.

There is another aspect that has not been explictly stated. Specification of the line integral does not uniquely define the E field. But if your world is cylindrically symmetric, then so is the E field.

 

[Adesh]

But if your world is cylindrically symmetric, then so is the E field.

All right! I understand that going a distance of $r$ from any point on $z$ axis in any direction would give me the same magnitude of $\mathbf B$ field and even the direction, have I understood you correctly?

 

[hutchphd]

The symmetry demands the same (rotated) vector field and the curl (or closed line integral) tells you the size of the "circular"(tangent) part.

 

[Adesh]

The symmetry demands the same (rotated) vector field

Sir, why symmetry demands that? I mean why going a distance $r$ in any direction from any point on $z$ axis (of course radially outward) will result in the same magnitude of electric field. I request you to please explain me that “demand” of symmetry.

 

[hutchphd]

There would need to be an external reason for the asymmetry. The laws themselves provide no preference. Only the charges/currents or boundary values can provide this and, by stipulation they do not. (Conversely if the solution is asymmetric it will show a charge or boundary asymmetry) QED