# I Why cant thelectric field be in circumferential direction?

#### garylau

Sorry

in this question i have several things to ask
1.why cant the E field in the circumferential direction?
2.How can i find the direction of B field

3.according to the formula of curl E=-dB/dt and the curlB=u J
did they share the same property so that the direction of E is circumferential (just like B dl=u i which is ampere law so that the direction B is circumferential) and dB/dt is going straight(just the the current is going straight which is enclosed inside the boundary)?

thank

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#### Simon Bridge

Homework Helper
1. I don't know what you mean by "circumferential".
Clearly it is possible to set up an electric field that goes in a loop: it's called an electric circuit.
But you cannot have a circular field line because the line has to start and end at different potentials.

2. You use a small magnetised needle balanced on a pivot.

3. The similarity in the equations shows that there is a similarity in the equations ...
Note, they should be:
$\nabla\times\vec E = -\partial_t\vec B$ and
$\nabla\times\vec B = \mu_0\left( \vec J +\epsilon_0\partial_t \vec E \right)$
... if $\vec J = 0$ the two equations look even more alike ;)

Check what "curl" means:
https://en.wikipedia.org/wiki/Curl_(mathematics)

#### garylau

1. I don't know what you mean by "circumferential".
Clearly it is possible to set up an electric field that goes in a loop: it's called an electric circuit.
But you cannot have a circular field line because the line has to start and end at different potentials.

2. You use a small magnetised needle balanced on a pivot.

3. The similarity in the equations shows that there is a similarity in the equations ...
Note, they should be:
$\nabla\times\vec E = -\partial_t\vec B$ and
$\nabla\times\vec B = \mu_0\left( \vec J +\epsilon_0\partial_t \vec E \right)$
... if $\vec J = 0$ the two equations look even more alike ;)

Check what "curl" means:
https://en.wikipedia.org/wiki/Curl_(mathematics)

you can see this picture clearly
the solution states:The magnetic field is “circumferential” in the quasistatic approximation."

.....but i dont know how did the solution concluded "Thus the direction of the electric field is longitudinal" from the simiarlty between the relation between the electric and magnetic fields?

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#### Cutter Ketch

What breaks the symmetry between the magnetic and the electric is the lack of magnetic monopoles: north and south magnetic charges that emanate magnetic field in the way positive and negative electric charges emanate electric field. If you had a magnetic wire carrying magnetic monopoles in a straight line they would generate a circumferential electric field in the same way an electric current generates a circumferential magnetic field.

#### garylau

What breaks the symmetry between the magnetic and the electric is the lack of magnetic monopoles: north and south magnetic charges that emanate magnetic field in the way positive and negative electric charges emanate electric field. If you had a magnetic wire carrying magnetic monopoles in a straight line they would generate a circumferential electric field in the same way an electric current generates a circumferential magnetic field.
What breaks the symmetry between the magnetic and the electric is the lack of magnetic monopoles: north and south magnetic charges that emanate magnetic field in the way positive and negative electric charges emanate electric field. If you had a magnetic wire carrying magnetic monopoles in a straight line they would generate a circumferential electric field in the same way an electric current generates a circumferential magnetic field.
But Why did the Electric field outside vanished in this case?

#### garylau

1. I don't know what you mean by "circumferential".
Clearly it is possible to set up an electric field that goes in a loop: it's called an electric circuit.
But you cannot have a circular field line because the line has to start and end at different potentials.

2. You use a small magnetised needle balanced on a pivot.

3. The similarity in the equations shows that there is a similarity in the equations ...
Note, they should be:
$\nabla\times\vec E = -\partial_t\vec B$ and
$\nabla\times\vec B = \mu_0\left( \vec J +\epsilon_0\partial_t \vec E \right)$
... if $\vec J = 0$ the two equations look even more alike ;)

Check what "curl" means:
https://en.wikipedia.org/wiki/Curl_(mathematics)
But Why did the Electric field outside vanished in this case?

#### Simon Bridge

Homework Helper
In what case? Please be specific.

#### garylau

In what case? Please be specific.
See the statement inside the red circle.

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#### Simon Bridge

Homework Helper
Oh so the question had nothing to do with the quoted text? Fine.
To see why the electric field vanishes, try doing the problem.

#### garylau

Oh so the question had nothing to do with the quoted text? Fine.
To see why the electric field vanishes, try doing the problem.
because the total current inside the object is 0?

#### mfb

Mentor
That contributes to it, yes, but the main point is the absence of magnetic charges and the fact that the problem assumes a static situation.

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