Question

Cathode rays of velocity $ {{10}^{6}}\,m/s $ describe an approximate circular path of radius $1 \,m$ in an electric field $300 \,V/cm$. 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 $F=e E\,\,\,...$ (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
$e E=\frac{m v^{2}}{r}$
$\Rightarrow r=\frac{m v^{2}}{e E}=\frac{m\left(10^{6}\right)^{2}}{e(300)}\,\,\,...$(iii)
When velocity is doubled same circular path is followed, hence radius is same
$r=\frac{m\left(2 \times 10^{6}\right)^{2}}{e E} \,\,\,...$(iv)
Equating Eqs. (iii) and (iv), we get
$\frac{m \times\left(10^{6}\right)^{2}}{300 e}=\frac{m \times\left(2 \times 10^{6}\right)^{2}}{e E}$
$\Rightarrow E=300 \times 4=1200 \,V / cm$
Note: Since, $E \propto u^{2}$ and $v$ is doubled much larger field will be required.