Question:
The equation that represents magnetic field of a plane electromagnetic wave which is propagating along \(x\)-direction with wavelength \(10\,\text{mm}\) and maximum electric field \(60\,\text{V m}^{-1}\) in \(y\)-direction is \((c=\text{speed of light})\):
The equation that represents magnetic field of a plane electromagnetic wave which is propagating along \(x\)-direction with wavelength \(10\,\text{mm}\) and maximum electric field \(60\,\text{V m}^{-1}\) in \(y\)-direction is \((c=\text{speed of light})\):
Show Hint
In an electromagnetic wave, \(\vec{E}\), \(\vec{B}\), and direction of propagation are mutually perpendicular. Also,
\[
E_0=cB_0
\]
and
\[
k=\frac{2\pi}{\lambda}.
\]
Updated On: Jun 26, 2026
- \((6\times 10^{-7})\sin[0.2\pi(ct-x)]\hat{k}\,\text{tesla}\)
- \((2\times 10^{-7})\sin[200\pi(ct-x)]\hat{k}\,\text{tesla}\)
- \((2\times 10^{-7})\sin[200\pi(ct-x)]\hat{i}\,\text{tesla}\)
- \((6\times 10^{-7})\sin[0.2\pi(ct-x)]\hat{i}\,\text{tesla}\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Identify the direction of magnetic field.
The electromagnetic wave is propagating along \(x\)-direction.
The electric field is along \(y\)-direction.
For an electromagnetic wave, the electric field, magnetic field, and direction of propagation are mutually perpendicular.
Also, \[ \vec{E}\times \vec{B} \] gives the direction of propagation.
Since propagation is along \(\hat{i}\) and electric field is along \(\hat{j}\), magnetic field must be along \(\hat{k}\), because \[ \hat{j}\times \hat{k}=\hat{i} \] Thus, magnetic field is along \[ \hat{k} \]
Step 2: Find the maximum magnetic field.
For an electromagnetic wave, \[ E_0=cB_0 \] Therefore, \[ B_0=\frac{E_0}{c} \] Given, \[ E_0=60\,\text{V m}^{-1} \] and \[ c=3\times 10^8\,\text{m s}^{-1} \] So, \[ B_0=\frac{60}{3\times 10^8} \] \[ B_0=20\times 10^{-8} \] \[ B_0=2\times 10^{-7}\,\text{T} \]
Step 3: Find the wave number.
Given wavelength is \[ \lambda=10\,\text{mm} \] Since, \[ 1\,\text{mm}=10^{-3}\,\text{m} \] Therefore, \[ \lambda=10\times 10^{-3}\,\text{m} \] \[ \lambda=10^{-2}\,\text{m} \] Wave number is \[ k=\frac{2\pi}{\lambda} \] \[ k=\frac{2\pi}{10^{-2}} \] \[ k=200\pi\,\text{m}^{-1} \]
Step 4: Write the magnetic field equation.
For a wave travelling along positive \(x\)-direction, the phase can be written as \[ k(ct-x) \] Therefore, \[ \vec{B}=B_0\sin[k(ct-x)]\hat{k} \] Substituting the values, \[ \vec{B}=(2\times 10^{-7})\sin[200\pi(ct-x)]\hat{k}\,\text{tesla} \]
Step 5: Final conclusion.
Therefore, the required magnetic field equation is \[ \boxed{(2\times 10^{-7})\sin[200\pi(ct-x)]\hat{k}\,\text{tesla}} \]
The electromagnetic wave is propagating along \(x\)-direction.
The electric field is along \(y\)-direction.
For an electromagnetic wave, the electric field, magnetic field, and direction of propagation are mutually perpendicular.
Also, \[ \vec{E}\times \vec{B} \] gives the direction of propagation.
Since propagation is along \(\hat{i}\) and electric field is along \(\hat{j}\), magnetic field must be along \(\hat{k}\), because \[ \hat{j}\times \hat{k}=\hat{i} \] Thus, magnetic field is along \[ \hat{k} \]
Step 2: Find the maximum magnetic field.
For an electromagnetic wave, \[ E_0=cB_0 \] Therefore, \[ B_0=\frac{E_0}{c} \] Given, \[ E_0=60\,\text{V m}^{-1} \] and \[ c=3\times 10^8\,\text{m s}^{-1} \] So, \[ B_0=\frac{60}{3\times 10^8} \] \[ B_0=20\times 10^{-8} \] \[ B_0=2\times 10^{-7}\,\text{T} \]
Step 3: Find the wave number.
Given wavelength is \[ \lambda=10\,\text{mm} \] Since, \[ 1\,\text{mm}=10^{-3}\,\text{m} \] Therefore, \[ \lambda=10\times 10^{-3}\,\text{m} \] \[ \lambda=10^{-2}\,\text{m} \] Wave number is \[ k=\frac{2\pi}{\lambda} \] \[ k=\frac{2\pi}{10^{-2}} \] \[ k=200\pi\,\text{m}^{-1} \]
Step 4: Write the magnetic field equation.
For a wave travelling along positive \(x\)-direction, the phase can be written as \[ k(ct-x) \] Therefore, \[ \vec{B}=B_0\sin[k(ct-x)]\hat{k} \] Substituting the values, \[ \vec{B}=(2\times 10^{-7})\sin[200\pi(ct-x)]\hat{k}\,\text{tesla} \]
Step 5: Final conclusion.
Therefore, the required magnetic field equation is \[ \boxed{(2\times 10^{-7})\sin[200\pi(ct-x)]\hat{k}\,\text{tesla}} \]
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 Electromagnetism Questions
- A charged particle moving along a straight-line path enters a uniform magnetic field of \( 4 \) mT at right angles to the direction of the magnetic field. If the specific charge of the charged particle is \( 8 \times 10^7 \) C/kg, the angular velocity of the particle in the magnetic field is:
- The current passing through a coil of 120 turns and inductance \( 40 \) mH is \( 30 \) mA. The magnetic flux linked with the coil is:
- To generate the high magnetic fields, the synchrotron uses
- The relation between permittivity of free space, the permeability of free space and speed of light is:
- A total charge \(q\) is uniformly distributed over a circular ring of radius \(R\). The ring is rotated about its axis with angular velocity \(\omega\). Find the magnetic dipole moment of the ring.
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