Question

A particle of mass m having a charge q enters into a circular region of radius R with velocity v directed towards the centre. The strength of magnetic field is B. Find the deviation in the path of the particle.

• A. $\pi -ta{n}^{-1}\left(\frac{mv}{\frac{qBR}{2}}\right)$
• B. $\pi -2ta{n}^{-1}\left(\frac{mv}{qBR}\right)$
• C. $\pi +ta{n}^{-1}\left(\frac{mv}{qB{R}^2}\right)$
• D. $\pi +2ta{n}^{-1}\left(\frac{mv}{qBR}\right)$

Answer 1

The correct option is B $\pi -2ta{n}^{-1}\left(\frac{mv}{qBR}\right)$


In $\Delta OBC$, $tan\alpha =\frac{r}{R}$
$\Rightarrow \alpha =ta{n}^{-1}\left(\frac{r}{R}\right)$
$=ta{n}^{-1}\left(\frac{mv}{qBR}\right)$
Now, deviation, $\delta =\pi -2\alpha$
$=\pi -2ta{n}^{-1}\left(\frac{mv}{qBR}\right)$