logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  EAMCET  >>  Physics
0 votes

an electron beam travels with a velocity of $1.6 \times 10^{7}\;ms^{-1}$ perpendicularly to magnetic field of intensity 0.1 T. The radius of the path of electron beam $(m_e=9 \times 10^{-31}kg)$

\[\begin {array} {1 1} (a)\;9 \times 10^{-5}m & \quad (b)\;9 \times 10^{-2}m \\ (c)\;9 \times 10^{-4}m & \quad (d)\;9 \times 10^{-3}m \end {array}\]

Can you answer this question?
 
 

1 Answer

0 votes
$(d)\;9 \times 10^{-3}m $
answered Nov 7, 2013 by pady_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
...