Question

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

Answer 1

According to the given criteria,

Focal length of the convex lens, $f=12cm$

Uniform magnetic field, $B=1.2T$

Charge on the particle, $q=2.0\times {10}^{-3}C$

Mass of the particle, $m=2.0\times {10}^{-6}kg$

Speed of the particle, $v=4.8m/s$

The distance of the particle from the lens, object distance, $u=-18cm$

As particle is projected perpendicular to the plane.

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

We know,

$r=\frac{mv}{qB}⟹r=\frac{2.0\times {10}^{-6}\times 4.8}{2.0\times {10}^{-3}\times 1.2}⟹r=4\times {10}^{-3}m=0.4cm$

Now applying the lens formula, we get

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}⟹\frac{1}{12}=\frac{1}{v}-\frac{1}{(-18)}⟹\frac{1}{v}=\frac{1}{12}-\frac{1}{18}⟹\frac{1}{v}=\frac{3-2}{36}⟹v=36cm$

Let the radius of the circular path of image be $r^{\prime }$

So magnification, $m=-\frac{v}{u}=\frac{r^{\prime }}{r}⟹-\frac{36}{(-18)}=\frac{r^{\prime }}{0.4}⟹r^{\prime }=\frac{36\times 0.4}{18}⟹r^{\prime }=0.8cm$

So the radius of the circular path in which the image movesis 0.8 cm.