question_answer

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field \[\vec{B}=(1.5\times {{10}^{-3}}T)\hat{k}\]at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is :

(electron's charge \[=1.6\times {{10}^{19}}C,\] mass of electron \[=9.1\times {{10}^{31}}\text{ }kg\])

[JEE Main 12-4-2019 Afternoon]

A) 12.87 cm

B) 1.22 cm

C) 11.65 cm                    

D) 2.25 cm

Correct Answer: A

Solution :

\[R=\frac{m\text{v}}{qB}\]\[=\frac{\sqrt{2m(K.E.)}}{qB}\] \[R=\frac{\sqrt{2\times 9.1\times {{10}^{-31}}\times (100\times 1.6\times {{10}^{-19}})}}{1.6\times {{10}^{-19}}\times 1.5\times {{10}^{-3}}}\] \[R=2.248cm\] \[\sin \theta =\frac{2}{2.248}\] \[\tan \theta =\frac{QU}{TU}\] \[\frac{2}{1.026}=\frac{QU}{6}\] \[QU=11.69\] \[PU=R(1-cos\theta )\] \[=1.22\] \[d=QU+PU\]