Question

Cathode rays of velocity ${10}^{6}m/s$ describe an approximate circular path of radius $1m$ in an electric field $300V/cm$. If the velocity of the cathode rays are doubled. The value of electric field so that the rays describe the same circular path, will be

• A. 2400 V/cm
• B. 600 V/cm
• C. 1200 V/cm
• D. 12000 V/cm

Answer 1

Answer: (C)

The electric field acting on the particles provides the necessary centripetal force. Cathode rays are composed of electrons, when they
move in electric field a force $F=eE...$ (і)
acts on them, this provides the necessary centripetal force to the particles.
$\therefore F=\frac{m{v}^2}{r}...$ (ii)
From Eqs. (i) and (ii), we get
$eE=\frac{m{v}^2}{r}$
$\Rightarrow r=\frac{m{v}^2}{eE}=\frac{m{\left(10^6\right)}^2}{e(300)}...$(iii)
When velocity is doubled same circular path is followed, hence radius is same
$r=\frac{m{\left(2\times 10^6\right)}^2}{eE}...$(iv)
Equating Eqs. (iii) and (iv), we get
$\frac{m\times {\left(10^6\right)}^2}{300e}=\frac{m\times {\left(2\times 10^6\right)}^2}{eE}$
$\Rightarrow E=300\times 4=1200V/cm$
Note: Since, $E\propto u^2$ and $v$ is doubled much larger field will be required.