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# A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 10^{8}ms^{–1}. It is then injected perpendicularly into a magnetic filed of strength 0.2 T. Find the radius of the circle described by it.

Answer :

Given-

Applied potential difference to the charged particle , *V* = 12 kV = 12 × 10^{3} V

Speed acquire by the charged particle, *v* =1.0 × 10^{6} m s^{−1}

Perpendicular Magnetic field *B* = 0.2 T

We know

The kinetic energy acquired by the particle is produced due to

applied potential difference, hence

and

where,

*m* is the mass of a proton

v= velocity of the particle

B = magnetic force

q= charge on the particle

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