Question

Cathode rays of velocity $10^{6}m{s}^{-1}$ describe an approximately circular path of radius 1 m in an electric field $300V{m}^{-1}.$ 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. $2400V{m}^{-1}$
• B. $600V{m}^{-1}$
• C. $1200V{m}^{-1}$
• D. $12000V{m}^{-1}$

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}