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Magnetic Field — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMagnetic Field5 Marks
Question
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 with 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 $4.8 ms^{-1}$​​​​​​​. 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 moves along a circle and find the radius of that circle.
✓

Answer



The object will make a circular path, perpendicular to the plance of paper Let the radius of the object be r.
$\frac{\text{mv}^2}{\text{r}}=\text{qvB}\Rightarrow\text{r}=\frac{\text{mV}}{\text{qB}}$
Here object distance K = 18cm.
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ (lens eqn.) $\Rightarrow\frac{1}{\text{v}}-\Big(\frac{1}{-18}\Big)=\frac{1}{12}\Rightarrow\text{v}=36\text{cm}.$
Let the radius of the circular path of image = r'
So magnification $=\frac{\text{v}}{\text{u}}=\frac{\text{r}'}{\text{r}}\Big(\text{magnetic path}=\frac{\text{image height}}{\text{object height}}\Big)$
$\Rightarrow\text{r}'=\frac{\text{v}}{\text{u}}\text{r}$
$\Rightarrow\text{r}'=\frac{36}{18}\times4=8\text{cm}$
Hence radius of the circular path in which the image moves is 8cm

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