In a vacuum chamber, a particle of charge 1 C and mass 1 mg is projected with a velocity ( i + 2 j ) ms ⁻¹ from the XZ plane at time t = 0 in an electric field of 1 i Vm ⁻¹ . At t = 0.2 s, the electric field is switched off and a magnetic field of 6 j T is switched on. The acceleration due to gravity is -10 j ms ⁻² . Correct option(s) is/are:

Step-by-step solution

For the time interval 0 t 0.2 s, the particle is under the influence of the electric field and gravity. The charge-to-mass ratio of the particle is q m = 10⁻⁶ 10⁻⁶ = 1 C kg ⁻¹ . The acceleration of the particle is: a = q E m + g = 1( i ) - 10 j = i - 10 j ms ⁻² Given the initial velocity v ₀ = i + 2 j ms ⁻¹ and initial vertical position y₀ = 0 (since it is projected from the XZ plane), we can find the velocity and position at t = 0.2 s. Velocity at t = 0.2 s: v (0.2) = v ₀ + a t = ( i + 2 j ) + ( i - 10 j )(0.2) =