Cathode rays of velocity 10 ^ 6 ^ -1 describe an approximately circular path of radius 1 m in an electric field 300 Vm ^ - 1 . If the velocity of cathode rays are doubled. The value of electric field so that the rays

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Correct answer: C. 1200 Vm ^ - 1

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

F = eE

...(i)

Acts on them, this provides the necessary centripetal force to the particles

F = \dfrac{mv^{2}}{r}

...(ii)

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

eE = \dfrac{mv^{2}}{r}

⇒ r = \dfrac{mv^{2}}{eE} = \dfrac{m{\left({10}^{6}\right)}^{2}}{e ( 300 )}

...(iii)

When velocity is doubled same circular path is followed, hence radius is same

r = \dfrac{m{\left(2 \times {10}^{6}\right)}^{2}}{eE}

...(iv)

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

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

⇒ E = 300 \times 4 = 1200

V

cm^{- 1}

Cathode rays of velocity 10 ^ 6 ^ -1 describe an approximately...

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