question_answer

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 \[v=\text{(2}\alpha d\text{)}\,(B),\] 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 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