A proton is moving with a uniform velocity of `10^(6)m.s^(-1)` along the Y-axis, under the joint action of a magnetic field along Z-axis and an electric field of magnitude `2xx10^(4)V.m^(-1)` along the negative X-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly (given: `(e)/(m)` ratio for proton `~~10^(8)C.kg^(-1)`)

Options

A. 0.5m
B. 0.2m
C. 0.1m
D. 0.05m

Answer

A

Explanation

Given `vec(v)= 10^(6) hat(j) ms^(-1)`,
`vec(E )= -2 xx 10^(4) hat(i)V m^(-1), vec(B)= B hat(k), (e )/(m)` (for proton) `= 10^(8) C kg^(-1)`
Force on Proton, `vec(F)= q (vec(E )+ vec(v) xx vec(B))`
`=e (-2 xx 10^(4) + 10^(6)B) hat(i)`
`:.` For undeflected proton, force, `vec(F)= 0` and magnetic field, `B= (2 xx 10^(4))/(10^(6))= 2 xx 10^(-2)T`
Again, radius of the circular path, `r= (m)/(e ) (v)/(B)`
`rArr r= (1)/(10^(8)) xx (10^(6))/(2 xx 10^(-2)) = 0.5m`