Question:
The magnetic field in a plane electromagnetic wave is given
\[
\vec{B}=(3\times10^{-7}\,\text{T})\sin(3\times10^4x+9\times10^{12}t)\,\hat{j}
\]
The electric field of this wave is given as
The magnetic field in a plane electromagnetic wave is given
\[
\vec{B}=(3\times10^{-7}\,\text{T})\sin(3\times10^4x+9\times10^{12}t)\,\hat{j}
\]
The electric field of this wave is given as
Show Hint
In an electromagnetic wave,
\[
E_0=cB_0
\]
and the vectors \(\vec{E}\), \(\vec{B}\), and direction of propagation are mutually perpendicular following the right-hand rule.
Updated On: Jun 22, 2026
- \(90\sin(3\times10^4x+9\times10^{12}t)\,\hat{i}\ \text{V m}^{-1}\)
- \(90\sin(3\times10^4x+9\times10^{12}t)\,\hat{k}\ \text{V m}^{-1}\)
- \(45\sin(3\times10^4x+9\times10^{12}t)\,\hat{i}\ \text{V m}^{-1}\)
- \(45\sin(3\times10^4x+9\times10^{12}t)\,\hat{k}\ \text{V m}^{-1}\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Use the relation between electric and magnetic fields.
For an electromagnetic wave, \[ E_0=cB_0 \] where, \[ c=3\times10^8\,\text{m s}^{-1} \] is the speed of light.
Given magnetic field amplitude, \[ B_0=3\times10^{-7}\,\text{T} \] Therefore, \[ E_0=(3\times10^8)(3\times10^{-7}) \] \[ E_0=9\times10^1 \] \[ E_0=90\,\text{V m}^{-1} \]
Step 2: Determine the direction of propagation.
The wave equation is \[ \sin(kx+\omega t) \] This represents propagation along the negative \(x\)-direction.
Thus, direction of propagation is \[ -\hat{i} \]
Step 3: Determine the direction of electric field.
For electromagnetic waves, \[ \vec{E}\times\vec{B} \] gives the direction of propagation.
Given, \[ \vec{B}\parallel \hat{j} \] We require \[ \vec{E}\times\hat{j}=-\hat{i} \] Using vector products, \[ \hat{k}\times\hat{j}=-\hat{i} \] Hence, \[ \vec{E}\parallel \hat{k} \]
Step 4: Write the electric field expression.
Therefore, \[ \vec{E}=90\sin(3\times10^4x+9\times10^{12}t)\,\hat{k}\ \text{V m}^{-1} \]
Step 5: Final conclusion.
Hence, the electric field is \[ \boxed{ 90\sin(3\times10^4x+9\times10^{12}t)\,\hat{k}\ \text{V m}^{-1} } \]
For an electromagnetic wave, \[ E_0=cB_0 \] where, \[ c=3\times10^8\,\text{m s}^{-1} \] is the speed of light.
Given magnetic field amplitude, \[ B_0=3\times10^{-7}\,\text{T} \] Therefore, \[ E_0=(3\times10^8)(3\times10^{-7}) \] \[ E_0=9\times10^1 \] \[ E_0=90\,\text{V m}^{-1} \]
Step 2: Determine the direction of propagation.
The wave equation is \[ \sin(kx+\omega t) \] This represents propagation along the negative \(x\)-direction.
Thus, direction of propagation is \[ -\hat{i} \]
Step 3: Determine the direction of electric field.
For electromagnetic waves, \[ \vec{E}\times\vec{B} \] gives the direction of propagation.
Given, \[ \vec{B}\parallel \hat{j} \] We require \[ \vec{E}\times\hat{j}=-\hat{i} \] Using vector products, \[ \hat{k}\times\hat{j}=-\hat{i} \] Hence, \[ \vec{E}\parallel \hat{k} \]
Step 4: Write the electric field expression.
Therefore, \[ \vec{E}=90\sin(3\times10^4x+9\times10^{12}t)\,\hat{k}\ \text{V m}^{-1} \]
Step 5: Final conclusion.
Hence, the electric field is \[ \boxed{ 90\sin(3\times10^4x+9\times10^{12}t)\,\hat{k}\ \text{V m}^{-1} } \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Electromagnetic waves Questions
- The RMS value of the electric field of an electromagnetic wave emitted by a source is \( 660 \) N/C. The average energy density of the electromagnetic wave is:
- The average value of electric energy density in an electromagnetic wave is: [where \( E_0 \) is the peak value]
- If the electric field of an electromagnetic wave is increased by 100%, then the energy density of the wave increases by
- Electromagnetic waves of energy flux $ 75 \times 10^4 \, \text{W/m}^2 $ incidents normally on a surface of area $ 40 \, \text{cm}^2 $. If the surface absorbs the flux completely, the total momentum delivered to the surface in one second is: Options:
- If $E_o$ and $B_o$ are the magnitudes of the electric and magnetic fields respectively of an electromagnetic wave in vacuum, then among the following the correct relation is ($\mu_o$ -- permeability of free spa, $\varepsilon_o$ -- permittivity of free spa)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions