Find Magnetic Field from Electric Field Using Maxwell's Equations

[ADCooper]

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.

 

[TSny]

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

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.

 

[ADCooper]

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!