Find Magnetic Field from Electric Field Using Maxwell's Equations

AI Thread Summary
The discussion revolves around calculating the magnetic field from a given electric field using Maxwell's equations. The initial attempt yielded a magnetic field expression that diverged from the expected solution, leading to confusion. The key issue identified was in the integration process of the time derivative of the magnetic field, which was incorrectly handled. After receiving feedback, the participant realized the mistake in their calculations and was able to correct their approach. The resolution highlights the importance of careful integration in electromagnetic theory.
ADCooper
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Homework Statement


An electromagnetic wave has an electric field \mathbf{E} = E_0 \cos(kz-ωt) \hat{x}. Using Maxwell's equations, find the magnetic field.

Homework Equations


\mathbf{∇\times E} = \mathbf{\dot{B}}

The Attempt at a Solution



So this problem appears extremely simple, but other students have told me my answer is incorrect, and I can't figure out what is wrong with my math. I find the cross product, which results in the following equation for the time derivative of the magnetic field:

\mathbf{\dot{B}} = \hat{y}kE_0\sin(kz-ωt)

I now integrate both sides with respect to time. This is where my answer diverges from others, so I'll fully write out my steps:

\mathbf{B} = kE_0∫_0^t \sin(kz-ωt') dt'\hat{y}

I set u = kz-ωt', which means du = -ωdt'

Plugging this in, the new integral is:

\mathbf{B} = -\frac{kE_0}{ω}∫_{kz}^{kz-ωt}\sin(u)du

The result is then

\mathbf{B} = \frac{kE_0}{ω}[\cos(kz-ωt)-\cos(kz)]\hat{y}

However, every student I've talked to has told me that the correct answer should be

\mathbf{B} = \frac{kE_0}{ω} \cos(kz-ωt)\hat{y}

Is there something simple I'm missing? There's nothing else in the problem description I didn't write. The other answer looks more correct but I can't find any reason that mine is incorrect.
 
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ADCooper said:
\mathbf{\dot{B}} = \hat{y}kE_0\sin(kz-ωt)

I now integrate both sides with respect to time. This is where my answer diverges from others, so I'll fully write out my steps:

\mathbf{B} = kE_0∫_0^t \sin(kz-ωt') dt'\hat{y}

The left side of the second equation above is incorrect. Think again about what you get when you integrate the left side of the first equation from t' = 0 to t' = t.
 
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TSny said:
The left side of the second equation above is incorrect. Think again about what you get when you integrate the left side of the first equation from t' = 0 to t' = t.

Woops thanks for the help, figured it out!
 
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