Question

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

• A. $2400 \, V \, m^{- 1}$
• B. $600 \, V \, m^{- 1}$
• C. $1200 \, V \, m^{- 1}$
• D. $12000 \, V \, m^{- 1}$

Answer 1

Answer: (C)

Cathode rays are composed of electrons, when they move in electric field a force $$ F=e E \quad \ldots \text { (i) } $$ Acts on them, this provides the necessary centripetal force to the particles $$ F=\frac{m v^{2}}{r} \quad \ldots \text { (ii) } $$ From Eqs. (i) and (ii), we get $$ \begin{array}{l} e E=\frac{m v^{2}}{r} \\ \Rightarrow \quad r=\frac{m v^{2}}{e E}=\frac{m\left(10^{6}\right)^{2}}{e(300)} \quad \ldots \text { (iii) } \end{array} $$ When velocity is doubled same circular path is followed, hence radius is same $$ r=\frac{m\left(2 \times 10^{6}\right)^{2}}{e E} \quad \ldots \text { (iv) } $$ Equating Eqs. (iii) and (iv), we get $$ \begin{array}{l} m \times \frac{\left(10^{6}\right)^{2}}{300 e}=\frac{m \times\left(2 \times 10^{6}\right)^{2}}{e E} \\ \Rightarrow \quad E=300 \times 4=1200 V cm ^{-1} \end{array} $$