A) 0.5cm
B) 2.5cm
C) 5.0cm
D) 7.5cm
Correct Answer: C
Solution :
Key Idea: Force due to perpendicular magnetic field provides the required centripetal force to the proton to move on circular path. Let there be a uniform magnetic field \[\vec{B}\]perpendicular to plane of paper, directed downward. Let a particle of mass m, and carrying a change + q enters the field with velocity v, then magnetic force = centripetal force i.e., \[q\,vB=\frac{m{{v}^{2}}}{r}\] where r is radius of circular path. Therefore, \[r=\frac{mv}{qB}\]. Given \[{{v}_{1}}=v,\,{{r}_{1}}=2.5\,cm,\,{{v}_{2}}=2\,v\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{v}_{1}}}{{{v}_{2}}}\] or \[{{r}_{2}}=\frac{{{v}_{2}}}{{{v}_{1}}}{{r}_{1}}\] or \[{{r}_{2}}=\frac{2\,v}{v}\times 2.5=5\,cm\].
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