Question:
An electron is constrained to move along the \(y\)-axis with a speed of \(0.1\,c\) (\(c\) is the speed of light) in the presence of electromagnetic wave, whose electric field is
\(\vec{E}=30\,\hat{j}\,\sin\left(1.5\times10^{7}t-5\times10^{-2}x\right)\) V/m.
The maximum magnetic force experienced by the electron will be :
(given \(c=3\times10^{8}\ \mathrm{ms^{-1}}\) and electron charge \(=1.6\times10^{-19}\) C)
An electron is constrained to move along the \(y\)-axis with a speed of \(0.1\,c\) (\(c\) is the speed of light) in the presence of electromagnetic wave, whose electric field is
\(\vec{E}=30\,\hat{j}\,\sin\left(1.5\times10^{7}t-5\times10^{-2}x\right)\) V/m.
The maximum magnetic force experienced by the electron will be :
(given \(c=3\times10^{8}\ \mathrm{ms^{-1}}\) and electron charge \(=1.6\times10^{-19}\) C)
\(\vec{E}=30\,\hat{j}\,\sin\left(1.5\times10^{7}t-5\times10^{-2}x\right)\) V/m.
The maximum magnetic force experienced by the electron will be :
(given \(c=3\times10^{8}\ \mathrm{ms^{-1}}\) and electron charge \(=1.6\times10^{-19}\) C)
Show Hint
For an EM wave \(B_0=E_0/c\). The wave travels along \(x\), \(\vec B\) is along \(z\) and the electron moves along \(y\), so \(\vec v\perp\vec B\) and \(F_{max}=evB_0\).
Updated On: Nov 5, 2026
- \(4.8\times10^{-19}\ \mathrm{N}\)
- \(2.4\times10^{-18}\ \mathrm{N}\)
- \(3.2\times10^{-18}\ \mathrm{N}\)
- \(1.6\times10^{-18}\ \mathrm{N}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Amplitude of the electric field: \(E_0=30\) V/m. Check: \(\omega/k = 1.5\times10^{7}/(5\times10^{-2})=3\times10^{8}\) m/s \(=c\).
Magnetic field amplitude: \(B_0=\frac{E_0}{c}=\frac{30}{3\times10^{8}}=10^{-7}\) T.
The wave propagates along \(+x\) with \(\vec E\) along \(y\), hence \(\vec B\) is along \(z\). The electron velocity is along \(y\), perpendicular to \(\vec B\).
\(F_{max}=e\,v\,B_0=1.6\times10^{-19}\times(0.1\times3\times10^{8})\times10^{-7}=1.6\times10^{-19}\times3\times10^{7}\times10^{-7}=4.8\times10^{-19}\) N.
Answer: (1) \(4.8\times10^{-19}\ \mathrm{N}\)
Magnetic field amplitude: \(B_0=\frac{E_0}{c}=\frac{30}{3\times10^{8}}=10^{-7}\) T.
The wave propagates along \(+x\) with \(\vec E\) along \(y\), hence \(\vec B\) is along \(z\). The electron velocity is along \(y\), perpendicular to \(\vec B\).
\(F_{max}=e\,v\,B_0=1.6\times10^{-19}\times(0.1\times3\times10^{8})\times10^{-7}=1.6\times10^{-19}\times3\times10^{7}\times10^{-7}=4.8\times10^{-19}\) N.
Answer: (1) \(4.8\times10^{-19}\ \mathrm{N}\)
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Electromagnetic waves Questions
- A source produced electromagnetic wave of frequency 60MHz. Find the wavelength of this wave in air.
- An electromagnetic wave has electric field given by \(\overrightarrow E = (9.6 \hat j)sin \bigg[2π\; {30 × 10^6t -\frac{1}{10} x}\bigg]\), \(x\) and \(t\) are in SI units. The maximum magnetic field is
- The electric field of an electromagnetic wave in free space is represented as \( \vec{E} = E_0 \cos(\omega t - kz) \hat{i} \). The corresponding magnetic induction vector will be:
- An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity \(6 \times 10^8 \, \text{W/m}^2\) falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space = \(3 \times 10^8\) m/s) :
- A particle is moving in a straight line. The variation of position \( x \) as a function of time \( t \) is given as\[x = (t^3 - 6t^2 + 20t + 15) \, \text{m}.\]The velocity of the body when its acceleration becomes zero is:
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions