question_answer

In a certain region of space there exists a constant and uniform magnetic field of induction B. The width of the magnetic field is a. A charged particle having charge q, is projected perpendicular to B and along the width of the field. If deflection produced by the field perpendicular to the width is d, then the magnitude of the momentum of the particle is

A) \[\left( \frac{{{d}^{2}}+{{a}^{2}}}{2d} \right)qB\]

B) \[\frac{{{a}^{2}}}{2{{d}^{2}}}qB\]

C) \[\frac{4{{a}^{2}}}{(a+d)}qB\]           

D) \[\frac{({{a}^{2}}-{{d}^{2}})}{2d}qB\]

Correct Answer: A

Solution :

[a] \[d-R\,(1-\cos \theta )\Rightarrow R\cos \theta =R-d\] \[R\sin \theta =a\Rightarrow {{R}^{2}}={{R}^{2}}-2Rd+{{d}^{2}}{{a}^{2}}\] \[\Rightarrow \,\,\,R=\frac{{{d}^{2}}+{{a}^{2}}}{2d}\,\,\,\,\Rightarrow \,\,\,p=\frac{{{d}^{2}}+{{a}^{2}}}{2d}qB\]