A) 0.2 cm
B) 1.2 cm
C) 2.2 cm
D) 0.122 cm
Correct Answer: B
Solution :
\[r=\frac{mv\sin \theta }{qB}\] \[r=\frac{1.67\times {{10}^{-27}}\times 4\times {{10}^{5}}\times \left( \frac{\sqrt{3}}{2} \right)}{1.6\times {{10}^{-19}}\times 0.3}\] \[=0.012m\] \[r=1.2cm\]You need to login to perform this action.
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