Calculation of magnetic field with help of electric field and frequency of em wave
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∇ · E = 0 (Gauss) ∇ · B = 0 (no name) ∇ × E = − ∂B (Faraday)∂t ∇ × B = μ0ε0 ∂E (Ampère)∂t
These equations are first order, which usually means the mathematics should be easy (good!), but they're also coupled, which means it might be difficult (rats!). Let's separate them using this little trick. Take the curl of both sides of Faraday's and Ampère's laws. The left side of each equation is the curl of the curl, for which there is a special identity. The right side of each equation, on the other hand, is the curl of a time derivative. We'll switch it around into a time derivative of the curl.
∇ × E = −∂B ∂t∇ × (∇ × E) = ∇ × ⎛
⎝− ∂B⎞
⎠∂t∇(∇ · E) − ∇2E = −∂ (∇ × B)∂t∇ × B = μ0ε0∂E ∂t∇ × (∇ × B) = ∇ × ⎛
⎝μ0ε0 ∂E⎞
⎠∂t∇(∇ · B) − ∇2B = μ0ε0∂ (∇ × E)∂t
Now if you look carefully, you'll see that one term in each equation equals zero and the other can be replaced with a time derivative.
0 − ∇2E = − ∂⎛
⎝μ0ε0 ∂E⎞
⎠∂t∂t0 − ∇2B = μ0ε0 ∂⎛
⎝−∂B⎞
⎠∂t∂t
Let's clean it up a bit and see what we get.
∇2E = μ0ε0 ∂2 E∂t2 ∇2B = μ0ε0 ∂2 B∂t2
These equations are now decoupled (E and B have their own private equations), which certainly simplifies things, but in the process we've changed them from first to second order (notice all the squares). I know I said earlier that lower order implies easier to work with, but these second order equations aren't as difficult as they look. Raising the order has not made things more complicated, it's made things more interesting.
Here's one set of possible solutions.
E (x,t) = E0 sin [2π(ft − x + φ)] ĵλB (x,t) = B0 sin [2π(ft − x + φ)] k̂λ
This particular example is one dimensional, but there are two dimensional solutions as well — many of them. The interesting ones have electric and magnetic fields that change in time. These changes then propagate away at a finite speed. Such a solution is an electromagnetic wave.
Let's examine our possible solution in more detail. Find the second space and time derivatives of the electric field…
∇2E = − 4π2E0 sin [2π(ft − x + φ)] ĵλ2λ∂2E = − 4π2f2E0 sin [2π(ft − x + φ)] ĵ∂t2λ
and substitute them back into the second order partial differential equation.
∇2E = μ0ε0∂2 E∂t2
Work on the left side first.
These equations are first order, which usually means the mathematics should be easy (good!), but they're also coupled, which means it might be difficult (rats!). Let's separate them using this little trick. Take the curl of both sides of Faraday's and Ampère's laws. The left side of each equation is the curl of the curl, for which there is a special identity. The right side of each equation, on the other hand, is the curl of a time derivative. We'll switch it around into a time derivative of the curl.
∇ × E = −∂B ∂t∇ × (∇ × E) = ∇ × ⎛
⎝− ∂B⎞
⎠∂t∇(∇ · E) − ∇2E = −∂ (∇ × B)∂t∇ × B = μ0ε0∂E ∂t∇ × (∇ × B) = ∇ × ⎛
⎝μ0ε0 ∂E⎞
⎠∂t∇(∇ · B) − ∇2B = μ0ε0∂ (∇ × E)∂t
Now if you look carefully, you'll see that one term in each equation equals zero and the other can be replaced with a time derivative.
0 − ∇2E = − ∂⎛
⎝μ0ε0 ∂E⎞
⎠∂t∂t0 − ∇2B = μ0ε0 ∂⎛
⎝−∂B⎞
⎠∂t∂t
Let's clean it up a bit and see what we get.
∇2E = μ0ε0 ∂2 E∂t2 ∇2B = μ0ε0 ∂2 B∂t2
These equations are now decoupled (E and B have their own private equations), which certainly simplifies things, but in the process we've changed them from first to second order (notice all the squares). I know I said earlier that lower order implies easier to work with, but these second order equations aren't as difficult as they look. Raising the order has not made things more complicated, it's made things more interesting.
Here's one set of possible solutions.
E (x,t) = E0 sin [2π(ft − x + φ)] ĵλB (x,t) = B0 sin [2π(ft − x + φ)] k̂λ
This particular example is one dimensional, but there are two dimensional solutions as well — many of them. The interesting ones have electric and magnetic fields that change in time. These changes then propagate away at a finite speed. Such a solution is an electromagnetic wave.
Let's examine our possible solution in more detail. Find the second space and time derivatives of the electric field…
∇2E = − 4π2E0 sin [2π(ft − x + φ)] ĵλ2λ∂2E = − 4π2f2E0 sin [2π(ft − x + φ)] ĵ∂t2λ
and substitute them back into the second order partial differential equation.
∇2E = μ0ε0∂2 E∂t2
Work on the left side first.
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