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 he 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 :
[B]Electromagnetic force will provide the necessary centripetal force. |
\[eqv=\frac{m{{v}^{2}}}{r}\] |
\[r=\frac{m{{v}^{2}}}{qB}=\frac{v}{B\alpha }=\frac{(2\alpha d)(B)}{(B\alpha )}=2d\] |
i.e. the electron will move out after traveling on a semicircular path of radius r=2d. Hence |
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