A) \[0.071\text{ }mT\]
B) \[0.71\text{ }mT\]
C) \[71\text{ }mT\]
D) \[7.1\text{ }mT\]
Correct Answer: B
Solution :
[b] Magnetic Force \[F=qvB\] \[\therefore \vec{a}=\left( \frac{qvB}{m} \right)\] Perpendicular to velocity. \[\therefore \]Also \[v=\sqrt{\frac{2k}{m}}=\sqrt{\frac{2\times e\times {{10}^{6}}}{m}}\] \[\therefore a=\frac{qvB}{m}=\frac{eB}{m}\sqrt{\frac{2\times e\times {{10}^{6}}}{m}}\] \[\therefore {{10}^{12}}={{\left( \frac{1.6\times {{10}^{-19}}}{1.67\times {{10}^{-27}}} \right)}^{\frac{3}{2}}}.\sqrt{2}\times {{10}^{3}}B\] \[\therefore B\simeq \frac{1}{\sqrt{2}}\times {{10}^{-3}}T=0.71\,mT(approx)\]You need to login to perform this action.
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