An electron is constrained to move along the

Question:

An electron is constrained to move along the y-axis with a speed of $0.1 \mathrm{c}$ ( $\mathrm{c}$ is the speed of light) in the presence of electromagnetic wave, whose electric field is

$\overrightarrow{\mathrm{E}}=30 \hat{\mathrm{j}} \sin \left(1.5 \times 10^{7} \mathrm{t}-5 \times 10^{-2} \mathrm{x}\right) \mathrm{V} / \mathrm{m}$.

The maximum magnetic force experienced by the electron will be :

(given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ and electron charge $=1.6 \times 10^{-19} \mathrm{C}$ )

  1. $1.6 \times 10^{-19} \mathrm{~N}$

  2. $4.8 \times 10^{-19} \mathrm{~N}$

  3. $3.2 \times 10^{-18} \mathrm{~N}$

  4. $2.4 \times 10^{-18} \mathrm{~N}$


Correct Option: , 2

Solution:

$\Rightarrow \mathrm{E}=\overrightarrow{\mathrm{E}}=30 \hat{\mathrm{j}} \sin \left(1.5 \times 10^{7} \mathrm{t}-5 \times 10^{-2} \mathrm{x}\right) \mathrm{V} / \mathrm{m}$

$\Rightarrow \mathrm{B} \Rightarrow \mathrm{E} / \mathrm{V} \Rightarrow \frac{30}{1.5 \times 10^{7}} \times 5 \times 10^{-2}$

$\Rightarrow 10^{-7}$ Tesla

$\Rightarrow \mathrm{F}_{\text {mag }}=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=|\mathrm{qVB}|$

$=1.6 \times 10^{-19} \times 0.1 \times 3 \times 10^{8} \times 10^{-7}$

$=4.8 \times 10^{-19} \mathrm{~N}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now