The electric field associated with an electromagnetic wave travelling in vacuum is given by
\[
\vec E
=
E_0\sin(3y+4z+\omega t)\,\hat i
\]
where \(\omega\) is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are:
\[
[\text{Given: speed of light in vacuum }c=3\times10^8\ \text{m s}^{-1}]
\]
Show Hint
- The wave is travelling in \(-\dfrac15(3\hat j+4\hat k)\) direction.
- The magnitude of wave vector is \(5\ \mathrm{m^{-1}}\)
- The value of \(\omega\) is \(1.5\times10^9\ \mathrm{rad\,s^{-1}}\)
- The magnetic field associated with this wave is given by \[ \vec B= \frac{E_0}{c} \sin(3y+4z+\omega t)\, (4\hat j-3\hat k) \]
The Correct Option is A
Solution and Explanation
Given: \[ \vec E= E_0\sin(3y+4z+\omega t)\,\hat i \] Compare with standard form: \[ \sin(\vec k\cdot\vec r+\omega t) \] Hence: \[ \vec k=3\hat j+4\hat k \] Magnitude: \[ |\vec k| = \sqrt{3^2+4^2} \] \[ =5\ \text{m}^{-1} \] Therefore: \[ \boxed{\mathrm{(B)\ is\ correct}} \]
Step 2: Find direction of propagation.
For: \[ \sin(\vec k\cdot\vec r+\omega t) \] wave travels opposite to: \[ \vec k \] Thus direction: \[ -\frac{\vec k}{|\vec k|} = -\frac15(3\hat j+4\hat k) \] Therefore: \[ \boxed{\mathrm{(A)\ is\ correct}} \]
Step 3: Find angular frequency.
For electromagnetic waves: \[ \omega=ck \] Thus: \[ \omega = (3\times10^8)(5) \] \[ =1.5\times10^9\ \text{rad s}^{-1} \] Therefore: \[ \boxed{\mathrm{(C)\ is\ correct}} \]
Step 4: Find magnetic field direction.
Direction relation: \[ \vec E\times\vec B \] gives propagation direction. Here: \[ \vec E\parallel\hat i \] Propagation direction: \[ -\frac15(3\hat j+4\hat k) \] Thus: \[ \vec B \propto -(4\hat j-3\hat k) \] Hence: \[ \vec B= -\frac{E_0}{c} \sin(3y+4z+\omega t) (4\hat j-3\hat k) \] Given option misses the negative sign. Therefore: \[ \boxed{\mathrm{(D)\ is\ incorrect}} \]
Step 5: Identify correct statements.
Therefore: \[ \boxed{\mathrm{(A),\ (B)\ and\ (C)}} \]
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