Calculation of magnetic field from electric field

In summary, the conversation discussed solving for the electric and magnetic fields in an EM wave using Lagrange's formula. The equations for the fields were given, along with Maxwell's equations, and the person was able to solve the problem using Lagrange's formula instead of the traditional method. This method allowed for a simpler and more efficient solution.
  • #1
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Homework Statement


In this tute on EM waves, we were given the Electric Field

[tex] \textbf{E}=\textbf{E}_0\text{exp}(i(\textbf{k}\cdot\textbf{x} - \omega t))[/tex]

which after a fair bit of algebra yields the magnetic field

[tex] \textbf{B}=(\hat{\textbf{k}}\times\textbf{E})/c[/tex]

Similarly the inverse problem I had to solve, given the Magnetic Field

[tex] \textbf{B}=\textbf{B}_0\text{exp}(i(\textbf{k}\cdot\textbf{x} - \omega t))[/tex]

yields [tex] \textbf{E}=c\textbf{B}\times\hat{\textbf{k}}[/tex]

The tute also gives a hint that this can be solved in a few lines, without heavy algebra, using Lagrange's formula.

Homework Equations



Maxwell's Equations

Lagrange's formula: [tex]\textbf{a}\times(\textbf{b}\times\textbf{c}) = (\textbf{a}\cdot\textbf{c})\textbf{b} - (\textbf{a}\cdot\textbf{b})\textbf{c}[/tex]

The Attempt at a Solution



I can only solve the question the long winded way, and would like to know how it can be solved using this identity rather than equating many equations to solve coefficients!

EDIT: Forgot to mention only using simplified Maxwell's Equations, i.e. Gauss' = 0 and Ampere's has no J term
 
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  • #2
I've managed to solve this myself, thanks for the help :cool:

[tex]\textbf{B}=(\hat{\textbf{k}}\times\textbf{E})/c[/tex]
[tex]c\textbf{B}=(\hat{\textbf{k}}\times\textbf{E})[/tex]

Then using Lagrange Triple Product
[tex]\hat{\textbf{k}}\times(\hat{\textbf{k}}\times\textbf{E}) = (\hat{\textbf{k}}\cdot\textbf{E})\hat{\textbf{k}} - (\hat{\textbf{k}}\cdot\hat{\textbf{k}})\textbf{E}[/tex]
[tex]\hat{\textbf{k}}\times c\textbf{B} = - \textbf{E}[/tex]

Therefore
[tex]\textbf{E} = c\textbf{B} \times \hat{\textbf{k}}[/tex]
//as required
 
  • #3


I would first commend the student for their efforts in solving the problem using the traditional method. It is important to understand the fundamentals before using shortcuts or identities.

That being said, Lagrange's formula can indeed be used to simplify the calculation of the electric field from the given magnetic field. We can start by using Ampere's law, which states that the curl of the magnetic field is equal to the current density. Since there is no current density given in this problem, we can simplify Ampere's law to:

∇ x B = 0

Now, using Lagrange's formula, we can rewrite this as:

∇ x (B x k) = (B x k) · ∇ - (B · ∇)(k)

Since we are only considering the simplified version of Maxwell's equations, we know that the divergence of the magnetic field is equal to zero, i.e. ∇ · B = 0. Therefore, the first term on the right side of the equation becomes zero. We are left with:

∇ x (B x k) = - (B · ∇)(k)

Now, we can use the fact that the magnetic field is given by B = B0exp(i(k · x - ωt)). Using the dot product rule, we can rewrite the second term on the right side of the equation as:

(B · ∇)(k) = B0 exp(i(k · x - ωt)) · (∇k)

Since k is a unit vector, its gradient is equal to zero. Therefore, we are left with:

(B · ∇)(k) = 0

Plugging this back into our equation, we get:

∇ x (B x k) = 0

Using another identity, ∇ x (A x B) = (B · ∇)A - (A · ∇)B, we can rewrite this equation as:

(B · ∇)k - (k · ∇)B = 0

Since k is a constant vector, its gradient is equal to zero. Therefore, we are left with:

(B · ∇)k = 0

Now, we can use the fact that k is a unit vector, and therefore (k · ∇)k = 0. This leaves us with:

(B · ∇) = 0

Finally, using the fact that the electric field is given by E
 

1. How do you calculate the magnetic field from an electric field?

The magnetic field can be calculated using the equation B = mu0 * (I / (2 * pi * r)), where B is the magnetic field, mu0 is the permeability of free space, I is the current, and r is the distance from the source of the electric field. Alternatively, you can use the right-hand rule to determine the direction of the magnetic field.

2. What is the relationship between electric and magnetic fields?

Electric and magnetic fields are interrelated and are both components of the electromagnetic force. A changing electric field can produce a magnetic field, and a changing magnetic field can produce an electric field. This phenomenon is described by Maxwell's equations.

3. How does the strength of the electric field affect the strength of the magnetic field?

The strength of the electric field is directly proportional to the strength of the magnetic field. This means that an increase in the strength of the electric field will result in an increase in the strength of the magnetic field, and vice versa.

4. What factors affect the calculation of the magnetic field from an electric field?

The calculation of the magnetic field from an electric field is affected by several factors, including the distance from the source of the electric field, the strength of the current, and the permeability of the medium through which the fields are passing.

5. Can the magnetic field be calculated for all types of electric fields?

Yes, the magnetic field can be calculated for all types of electric fields, including those produced by moving charges, alternating currents, and static charges. However, the calculation method may vary depending on the specific type of electric field.

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