logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Waves
0 votes

A beam of light travelling along $ x$ - axis is described by the electric field $E_y = (600 V/m) \sin \omega (t - x/c)$ then maximum magnetic force on a charge $q = 2e$, moving along $y$ - axis with a speed of $3.0 \times 10^7m/s$ is $(e = 1.6 \times 10^{-19} C)$

$\begin {array} {1 1} (a)\;19.2 \times 10^{-17}N & \quad (b)\;1.92 \times 10^{-17}N \\ (c)\;0.192 N & \quad (d)\;\text{None of these} \end {array}$

 

Can you answer this question?
 
 

1 Answer

0 votes
Maximum magnetic field is given by $B_o = \large\frac{E_o}{c}$
Here, $E_o = 600\: V/m$
So, $B_o = \large\frac{600}{3 \times 10^8} $$= 2 \times 10^{-6}T$
Maximum magnetic force imposed on a given charge is
$F_m = qvB_o = 2evB_o = 2 \times 1.6 \times 10^{-19} \times 3 \times 10^7 \times 2 \times 10^{-6}$
$= 1.92 \times 10^{-17}N$
Ans : (b)
answered Feb 27, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...