Sol. The Lorentz force on the charged particle is

F =q( + × )

The electric force on the charged particle, F E = qE Z , which acts towards negative direction z in the direction of electric field.

The magnetic force on the charged particle,

F B = qv x B y

As velocity of charge is in +x-direction and magnetic field is along +y-direction. From right hand rule the magnetic force acts along positive z-direction.

The resultant force

F = F E + F B = q(E 2 + v x × B y )

= q[–102.4 × 10 3 + 1.28 × 10 6 × 8 × 10 –2 ]

= 0

During time t = 0 to t 1 = 5 × 10 –6 the resultant force on the particle is zero, it moves with uniform velocity v x . The position of the particle (X 1 , Y 1 , Z 1 ) after time t 1 is

X 1 = v x t 1 = (1.28 × 10 6 ) × (5 × 10 –6 ] = 6.4 m

When electric field is switched off, the particle circulates in xz-plane under the influence of magnetic field.

Radius R of circulation is

R = = = 1 m

Let the particle rotate by an angle θ = ω (t 2 – t 1 ). The arc length P 1 P 2 = R θ = v x (t 2 – t 1 ), as the particle circulates for

t 2 – t 1 = (7.45 – 5) × 10 –6 = 2.45 × 10 –6 s

θ = =

= 3.136 π radians.

The coordinates of the particle are

X 2 = X 1 + R sin θ

= X 1 + R sin π

= X 1

= 6.4 m

and Z 2 = R – R cos θ

= R – R cos π

= 2R

= 2 m

Note that θ = π implies that t 2 = T/2 where T is time period of circulation. We could have written the result directly.