A uniform electric field `vecE` in the y-direction and uniform magnetic field `vecB` in the x-direction exists in free space. A particle of mass m and carrying charge g is projected from the origin with speed `v_0` along the y axis. The speed of the particle as a function of its y coordinate will be

Options

A. `sqrt(v_(0)^(2)+(2qEy)/(m))`
B. `sqrt(v_(0)^(2)+(qEy)/(m))`
C. `sqrt(v_(0)^(2)+(4qEy)/(m))`
D. `v_(0)`

Answer

A

Explanation

Here, `vec E = E hat j , vec B = B hat i , vec v= v_0 hatj`
The Lorentz force acting on a charged particle of charge a and mass m is
` vec F = q(vecE + vecv xx vecB) = q(E hatj + v_0hatj xxB hat i) = q(E hat j -v_0Bhat k)`
`therefore a_y = (aE)/(m) ...(i) and a_z = -(qv_0B)/(m) ...(ii)`
From equation (i), we get
`(dv_y)/(dt) = (qE)/(m) , (dv_y)/(dy) (dy)/(dt) = (qE)/(m)`