Question

If a point sized object having charge 1 C and mass 1 g is projected with a velocity of `2hat i` m/s from a point (0, 2 cm, 0) in the region of a magnetic field of −0.1 `hat k` T which spreads in the first quadrant.

1. What will be the shape of the path followed by the given charged particle?
2. At what point will it cross the X-axis?
3. What will be the kinetic energy of the particle when it enters the fourth quadrant?

Answer 1

Given:

Charge (q) = 1 C

Mass (m) = 1 g = 10⁻³ kg

Velocity `(vecv) = 2hat i` m/s

Magnetic field `(vecB) = -0.1 hat k T`

Initial point = (0, 2 cm)

A. Shape of the path:

`vecF = q(vecv xx vecB)`

`vecv = +hati, vecB = -hatk`

`hati xx (-hatk) = +hatj`

Thus, the particle follows a circular path in a clockwise direction, with magnetic force acting as centripetal force.

B. The charged particle will not cross the X-axis.

Reason: The velocity of the particle is perpendicular to the uniform magnetic field; therefore, the magnetic force acts perpendicular to the velocity and produces uniform circular motion with radius

r = `(mvecv)/(qvecB)`

= `(10^-3 xx 2)/(1 xx 0.1)`

= 0.02 m

= 2 cm

Since the initial position is (0, 2 cm) and the magnetic force is directed upward, the center of the circular path lies at (0, 4 cm).

Thus, the entire circular trajectory lies in the region y ≥ 2 cm, and the particle never reaches the X-axis.

C. Magnetic force performs no work:

W = 0

So kinetic energy remains constant.

K = `1/2 mv^2`

= `1/2 xx 10^-3 xx 2^2`

= `1/2 xx 10^-3 xx 4`

= 2 × 10⁻³ J