A) \[{{M}_{1}}{{v}_{1}}<{{M}_{2}}{{v}_{2}}\]
B) \[{{M}_{1}}{{v}_{1}}>{{M}_{2}}{{v}_{2}}\]
C) \[{{M}_{1}}{{v}_{1}}={{M}_{2}}{{v}_{2}}\]
D) None of these
Correct Answer: B
Solution :
If R be the radius of a particle of mass M and charge q moving with velocity v perpendicular to a uniform magnetic field B given as then \[\frac{M{{V}^{2}}}{R}=qvB\] \[\Rightarrow \,\,\,\,\,MV=qRB\] \[\therefore \,\,\,\,\,\,\,For\,\,particle\,\,1,\,\,{{M}_{1}}{{v}_{1}}\,\,=\,\,q{{R}_{1}}B\] ... (i) \[and\,\,for\,\,particle\,\,2,\,\,{{M}_{2}}{{v}_{2}}\,=\,\,q{{R}_{2}}B\] ... (ii) On dividing Eq. (i) by Eq. (ii), we get \[\therefore \,\,\,\,\,\,\,\,\,\frac{{{M}_{1}}{{v}_{1}}}{{{M}_{2}}{{v}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\] According to the figure \[{{R}_{1}}>{{R}_{2}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,{{M}_{1}}{{v}_{1}}>{{M}_{2}}{{v}_{2}}\]You need to login to perform this action.
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