Question

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

Answer 1

The correct option is C $1200Vc{m}^{-1}$

Cathode rays are composed of electrons, when they move in electric filed a force

$F=eE$ ...(i)

its acts on them and provides the necessary centripetal force the particles

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

From Eqs. (i) and (ii) we get

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

$r=\frac{m{v}^2}{eE}=\frac{m({10}^6{)}^2}{e\left(300\right)}$ ...(iii)

when, velocity is doubled same circular path is followed. Hence, radius is same

$r=\frac{m(2\times {10}^6{)}^2}{eE}$ ...(iv)

Equating Eqs. (iii) and (iv) we get

$\frac{m\times ({10}^6{)}^2}{e\left(300\right)}=\frac{m\times (2\times {10}^6{)}^2}{eE}$

$E=300\times 4=1200V/m$