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  • A particle of mass m and charge q, accelerated by a potential difference V enters a region of a uniform transverse magnetic field B. If d is the thickness of the region of B, the ang

A particle of mass m and charge q, accelerated by a potential difference V enters a region of a uniform transverse magnetic field B. If d is the thickness of the region of B, the angle q through which the particle deviates from the initial direction on leaving the region is given by

Option: 1

\mathrm{\sin \theta=B d\left(\frac{q}{2 m V}\right)^{1 / 2} }


Option: 2

\mathrm{\cos \theta=B d\left(\frac{q}{2 m V}\right)^{1 / 2} }


Option: 3

\mathrm{\tan \theta=B d\left(\frac{q}{2 m V}\right)^{1 / 2} }


Option: 4

\mathrm{\cot \theta=B d\left(\frac{q}{2 m V}\right)^{1 / 2}}


Answers (1)

best_answer

Refer to the below Figure. Let v be the velocity of the particle. Its kinetic energy is
\mathrm{\frac{1}{2} m v^2=q V \text { or } v=\left(\frac{2 q V}{m}\right)^{1 / 2} }...................(1)

The particle follows a circular path from A to B of radius r which is given by
\mathrm{ \frac{m v^2}{r}=q v B \text { or } r=\frac{m v}{q B} }............................(2)
Using (1) and (2), we have
\mathrm{ r=\frac{m}{q^B}\left(\frac{2 q V}{m}\right)^{1 / 2}=\frac{1}{B}\left(\frac{2 m V}{q}\right)^{1 / 2} }
In triangle BCD \mathrm{ \sin \theta=\frac{B D}{B C}=\frac{d}{r} } . Therefore, \mathrm{ \sin \theta=B d\left(\frac{q}{2 m V}\right)^{1 / 2} }, which is choice (a).

 

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Riya

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