$\begin {array} {1 1} (a)\;22.7\: m & \quad (b)\;2.27\: cm \\ (c)\;22.7\: cm & \quad (d)\;2.27\: m \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (c)

$v=5.2 \times 10^6m/s\: \: \: \: \: \:B= 1.3 \times 10^{-4}T$

So, Force exerted on the electron is : $F = e|v \times B| = evB\: \sin \theta $

$( \theta$=Angle between magnetic field and beam velocity)

So, $ \theta=90^{\circ}$

$F=evB$………………..(i)

The beam traces a circular path of radius, $r$.

It is the magnetic field, due to its bending nature, that provides the

centripetal force $(F=\large\frac{mv^2}{r})$ for the beam.

Hence, Eq. (i) reduces to

$evB = \large\frac{mv^2}{r}$

So, $r= \large\frac{mv}{eB} = \large\frac{v}{ \bigg[\bigg(\Large\frac{e}{m}\bigg) B\bigg]}$

$= \large\frac{5.2 \times 10^6 }{(1.76 \times 10^{11} \times 1.3 \times 10^{-4})}$

= 0.227 m = 22.7 cm

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...