Question

The figure shows a convex lens of focal length 12 cm lying in a uniform magnetic field B of 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:
$r=\frac{mv}{qB}$
$$\begin{gathered} r=\frac{2\times {10}^{-5}\times 4.8}{2\times {10}^{-3}\times 1.2} \\ r=0.04m=4cm \end{gathered}$$
Here, object distance, u = -18 cm
Using the lens equation
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f}, \\ \frac{1}{v}-\frac{1}{\left(-18\right)}=\frac{1}{12} \end{gathered}$$
Image distance, v = 36 cm.
Let the radius of the circular path of image be r'.
So, magnification:
$$\begin{gathered} \frac{v}{u}=\frac{r\text{'}}{r} \\ r\text{'}=\frac{v}{u}\times r \\ =8\mathrm{cm} \end{gathered}$$

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