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A proton moves in a circular path perpendicular to a magnetic field. If the intensity of the field is doubled and the radius of the circular path is constant, then the kinetic energy of the particle has

$\begin {array} {1 1} (a)\;Halved & \quad (b)\;Doubled \\ (c)\;Become\: quarter & \quad (d)\;Become\: four \: times \end {array}$

 

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$R = \large\frac{mV}{qB}$
$KE = \large\frac{1}{2} $$mV^2$
Ans : (d)
answered Feb 21, 2014 by thanvigandhi_1
 

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