Question

Cathode rays of velocity ${10}^{6}m{s}^{-1}$ describe an approximate circular path of radius 1m in an electric field of $400Vc{m}^{-1}.$ If the velocity of cathode rays is doubled, the value of the electric field needed so that the rays describe the same circular path is

• A. $1200Vc{m}^{-1}$
• B. $100Vc{m}^{-1}$
• C. $800Vc{m}^{-1}$
• D. $1600Vc{m}^{-1}$

Answer 1

Answer: (D)

$1600Vc{m}^{-1}$

Given, path of cathode rays is circular. So, electric force must be acting as centripetal force for the cathode rays and it is given by
$F_e=F_c$

qE = $m\frac{v^2}{r}$

we can conclude that $E\alpha v^2$ ,provided all the other quantities are constant.
Now, let $(E_i,V_i)and(E_f,V_f)$ are initial and final electric field strenthgs and velocities
Given, $V_f=2V_i$

solving $\frac{E_i}{E_f}=\frac{(2V_i{)}^2}{V_1}$ we get $E_f=4E_i$

As, $E_i=400Vc{m}^{-1}$
$E_f=1600Vc{m}^{-1}$

So, the answer is option (D).