A) \[2.9\,\,cm\]
B) \[3.9\,\,cm\]
C) \[2.35\,\,cm\]
D) \[2\,\,cm\]
Correct Answer: C
Solution :
Key Idea: Magnetic force provides the required centripetal force for motion on circular path. For an electron with charge q moving q with velocity \[v\], in a magnetic field B, inclined at angle \[\theta \], the force experienced is \[F=q\,v\,B\,\,\sin \,\theta \] ?(1) The electron follows a circular path, hence centripetal force is given by \[F=\frac{m{{v}^{2}}}{r}\] ?(2) Which is the result of magnetic force acting on it. Where \[r\] is radius of circular path. Equating Eqs. (1) and (2), we get \[q\,v\,B\,\sin \,\theta =\frac{m{{v}^{2}}}{r}\] Given, \[\theta ={{90}^{\circ }}\], therefore \[\sin {{90}^{\circ }}=1.\] \[\therefore \] \[qvB=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \] \[r=\frac{mv}{qB}\] \[\Rightarrow \] \[r=\frac{v}{\left( \frac{q}{m} \right)B}\] \[\left( \frac{q}{m}=\text{Specific charge} \right)\] Given, \[\frac{q}{m}=1.7\times {{10}^{11}}\,C/kg,\,v=6\times {{10}^{7}}\,m/s\] \[B=1.5\times {{10}^{-2}}\,\,T\] \[\therefore \] \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\] \[=2.35\times {{10}^{-2}}\,m\] \[=2.35\,cm\]
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