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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.
Solution
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.
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