(electron's charge \[=1.6\times {{10}^{19}}C,\] mass of electron \[=9.1\times {{10}^{31}}\text{ }kg\]) |
A) 12.87 cm
B) 1.22 cm
C) 11.65 cm
D) 2.25 cm
Correct Answer: A
Solution :
\[R=\frac{m\text{v}}{qB}\]\[=\frac{\sqrt{2m(K.E.)}}{qB}\] \[R=\frac{\sqrt{2\times 9.1\times {{10}^{-31}}\times (100\times 1.6\times {{10}^{-19}})}}{1.6\times {{10}^{-19}}\times 1.5\times {{10}^{-3}}}\] \[R=2.248cm\] \[\sin \theta =\frac{2}{2.248}\] \[\tan \theta =\frac{QU}{TU}\] \[\frac{2}{1.026}=\frac{QU}{6}\] \[QU=11.69\] \[PU=R(1-cos\theta )\] \[=1.22\] \[d=QU+PU\]You need to login to perform this action.
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