Answer to Question #304904 in Electric Circuits for Blue

Question

A 150-V battery is connected across two parallel metal plates of area 28.5 cm² and separation 8.20 mmm. A beam of alpha particles (charge +2ê , mass 6.64 x 10-²⁷ kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

Answer 1

Electric force $=$ Magnetic force

$F_e=F_B$

$qE=qvB\implies\>B=\frac{E}{v}$

$E=\frac{V_c}{d}$ Where $V_c=$ p.d across capacitor

$=\frac{150}{8.20×10^{-3}}$ $=18292.68$

Velocity, v, of alpha particles is given by

$qV_a=\frac{1}{2}mv^2$ Where $V_a$ is the accelerating potential

$\therefore\>\>V=\sqrt{\frac{2×2×1.6×10^{-19}×1.75×10^3}{6.64×10^{-27}}}$

$=\sqrt{1.687×10^{11}}=410700$

$B=\frac{18292.68}{410700}=4.454×10^{-2}$ T