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

Explanation

Initially the Lorentz force acting on the proton,
`vecF=vecF_(e)+vecF_(m)=vecE_(q)+qvecvxxvecB=-Eqhati+qvBhati`

Since the proton has no acceleration,
`:.vecF=0":.Eq=qvB`
or, `B=(E)/(v)=(2xx10^(4))/(10^(6))=2xx10^(-2)T`
When the electric field is switched off then the radius of the circular path,
`R=(mv)/(qB)=(10^(6))/(10^(8)xx2xx10^(-2))=0.5m`