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Question

A charged particle moves in a uniform magnetic field perpendicular to it, with a radius of curvature 4 cm. On passing through a metallic sheet, it loses half of its kinetic energy. Then, the radius of curvature of the particle is

A
2 cm
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B
4 cm
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C
8 cm
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D
22 cm
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Solution

The correct option is D 22 cm
Let the magnetic field be perpendicular to the plane (-z direction) and initial particle velocity be v along x direction. Then the particle will execute uniform circular motion in xy plane.


Radius of the circular path of the particle r=mvqB(1)
mv=p (momentum) =2mK
where K is the kinetic energy of the particle.
r=2mKqB

After passing through the metal sheet, let the new radius of curvature be r.
Then r=2mKqB(2)

(1) ÷ (2) rr=12
r=r2=42
=22 cm

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