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Moving Charges and Magnetism question

2015 · Q116
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Moving Charges and Magnetism question

2015 · Q116

NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be
  1. A
    1.5 MeV
  2. B
    1 MeV
  3. C
    4 MeV
  4. D
    0.5 MeV
View written solutionFree

Correct answer: B

As we know, F = qvB = mv2R{{m{v^2}} \over R}Rmv2​

∴\therefore∴ R=mvqB=2m(kE)qBR = {{mv} \over {qB}} = {{\sqrt {2m\left( {kE} \right)} } \over {qB}}R=qBmv​=qB2m(kE)​​

Since R is same so, KE∝q2mKE \propto {{{q^2}} \over m}KE∝mq2​

Therefore KE of α\alpha α particle

=q2m=(2)24=1 MeV = {{{q^2}} \over m} = {{{{\left( 2 \right)}^2}} \over 4} = 1\,MeV=mq2​=4(2)2​=1MeV

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