Question

The figure shows a convex lens of focal length 12 cm lying in a uniform magnetic field Bof magnitude 1.2 T parallel to its principal axis. A particle with charge 2.0 × 10⁻³ C  and mass 2.0 × 10⁻⁵ kg is projected perpendicular to the plane of the diagram with a speed of 4.8 m s⁻¹. The particle moves along a circle with its centre on the principal axis at a distance of 18 cm from the lens. Show that the image of the particle moves along a circle and find the radius of that circle.

Answer 1

Given:-

Focal length of the convex lens = 12 cm

Uniform magnetic field, B = 1.2 T

Charge of the particle, q = 2.0 × 10⁻³ C

and mass, m = 2.0 × 10⁻⁵ kg

Speed of the particle, v = 4.8 m s⁻¹

The distance of the particle from the lens = 18 cm

As per the question, the object is projected perpendicular to the plane of the paper.

Let the radius of the circle on which the object is moving be r.

We know:

`r = (mv)/(qB)`

`r = (2xx10^-5xx4.8)/(2xx10^-3xx1.2)`

`r = 0.04  m` = 4 cm

Here, object distance, u = -18 cm

Using the lens equation

`1/v - 1/u = 1/f`,

`1/v - 1/(-18) = 1/12`

Image distance, v = 36 cm.

Let the radius of the circular path of image be r'.

So, magnification:

`v/u = (r')/r`

`r'= v/u xx r`

= 8 cm

Therefore, the radius of the circular path in which the image moves is 8 cm.