If a charged particle of charge to mass ratio \[\frac{q}{m}=\alpha \] is entering in a magnetic field of strength B at a speed \[v=(2\,\alpha \,d)\,\,(B),\] then which of the following is correct: |
A) angle subtended by charged particle at the centre of circular path is \[2\pi .\]
B) the charge will move on a circular path and will come out from magnetic field at a distance 4d from the point of insertion.
C) the time for which particle will be in the magnetic field is \[\frac{2\pi }{\alpha B}.\]
D) the charged particle will subtend an angle of \[90{}^\circ \] at the centre of circular path
Correct Answer: B
Solution :
Electromagnetic force will provide the necessary centripetal force. |
\[eqv=\frac{m{{v}^{2}}}{r}\] |
\[\Rightarrow \] \[r=\frac{mv}{qB}=\frac{v}{B\,\alpha }=\frac{(2\,\alpha \,d)\,\,(B)}{(B\,\alpha )}=2\,d\] |
i.e. the electron will move out after travelling on a semicircular path of radius \[r=2d.\] |
Hence [B] |
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