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FInd a simple expression for the vector potential which will yield this field

  • Thread starter Nusc
  • Start date
  • #1
754
2
em

1. Homework Statement

In a certain region, the magnetic field as a linear function of time is given by

[tex]
B = Bo \frac{t}{/tau} z hat
[/tex]
Bo and tau constants.

A)FInd a simple expression for the vector potential which will yield this field.
B)Assuming the scalar potential is a constant, find E from the above result for the vector potential. Check that your answer is consistent with the diff form of faradays law

2. Homework Equations
[tex]
\nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}
[/tex]

[tex]
\mathbf{A} = \frac{\mu_0}{4 \pi} \int \frac{\mathbf{J}(\mathbf{rsingle-quote},t_r)}{\cal{R}} d\tausingle-quote
[/tex]

[tex]
t_r \equiv t - \frac{\cal{R}}{c}
[/tex]


3. The Attempt at a Solution

[tex]
B= \nabla \times A
[/tex]

We take the curl of B on the LHS and on the RHS we have
[tex]
\nabla \times (\nabla \times A )
[/tex]

Eventually we will get to this [tex]
\nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}
[/tex]


This is assuming were in the Lorentz Gauge, we don't know if we are. Or do we assume we are? We have to satisfy E = -delV -dA/dt but we don't know either of those quantities.
 

Answers and Replies

  • #2
754
2
say if

kt y-hat = K x xhat

What mathematical operation would I apply to show that
K = kt z-hat

I don't thin kthat is even relevent
 
Last edited:
  • #3
pam
455
1
A) Take curl(BXr)/2 and see what you get.
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,540
181
1. Homework Statement

In a certain region, the magnetic field as a linear function of time is given by

[tex]
B = Bo \frac{t}{/tau} z hat
[/tex]
Bo and tau constants.

A)FInd a simple expression for the vector potential which will yield this field.
B)Assuming the scalar potential is a constant, find E from the above result for the vector potential. Check that your answer is consistent with the diff form of faradays law

2. Homework Equations
[tex]
\nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}
[/tex]

[tex]
\mathbf{A} = \frac{\mu_0}{4 \pi} \int \frac{\mathbf{J}(\mathbf{rsingle-quote},t_r)}{\cal{R}} d\tausingle-quote
[/tex]

[tex]
t_r \equiv t - \frac{\cal{R}}{c}
[/tex]


3. The Attempt at a Solution

[tex]
B= \nabla \times A
[/tex]

We take the curl of B on the LHS and on the RHS we have
[tex]
\nabla \times (\nabla \times A )
[/tex]

Eventually we will get to this [tex]
\nabla^2 \mathbf{A} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0 \mathbf{J}
[/tex]


This is assuming were in the Lorentz Gauge, we don't know if we are. Or do we assume we are? We have to satisfy E = -delV -dA/dt but we don't know either of those quantities.
The simples way to go about this is to simply write explicitly [tex]
B= \nabla \times A
[/tex]
in cartesian coordinates.

You will get simple equations for A_x, A_y and A_z. It should be easy to pick functions A-x, A_y, A_z that satisfy those equations. You don't need anything else.
 
  • #5
754
2
The simples way to go about this is to simply write explicitly [tex]
B= \nabla \times A
[/tex]
in cartesian coordinates.

You will get simple equations for A_x, A_y and A_z. It should be easy to pick functions A-x, A_y, A_z that satisfy those equations. You don't need anything else.
Doing this would only yield

dA_y/dx - dA_x/dy = Bo(t/tau) since B has only z-components.

Solving this differential equation would still give unknown A_y A_x.

There should be an explicit way of doing this, no?
 
  • #6
979
1
Remember that the vector potential is not physically different even if you add a gradient of a function. So you have a certain freedom available to pick the solution.
 
  • #7
nrqed
Science Advisor
Homework Helper
Gold Member
3,540
181
Doing this would only yield

dA_y/dx - dA_x/dy = Bo(t/tau) since B has only z-components.

Solving this differential equation would still give unknown A_y A_x.

There should be an explicit way of doing this, no?
Of course A is defined up to gauge transformations only so many different A's give the same physics. This is why the question asks you to find a simple expression for A. Just pick the simplest A that yields the correct B

(don't forget to make sure that the other components of the equation also work)
 
  • #8
754
2
OKay problem solved. You can close this thread.
 

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