Question

An electron is accelerated through a potential difference of 3.5 kV and directed into a region between two parallel plates separated by 29 mm with a potential difference of 100 V between them. The electron is moving perpendicular to the electric field when it enters the region between the plates. What magnetic field is necessary perpendicular to both the electron path and the electric field so that the electron travels in a straight line

Answer 1

Answer: $0.985\times 10^{-4}\ T$

Explanation:

Given

Electron is accelerated 3.5 kV potential difference

Distance between plates d=29 mm

The potential difference between plates is V=100 V

here, the kinetic energy of an electron is acquired through a potential difference of 3.5 kV

$\Rightarrow \dfrac{1}{2}m_ev^2=e\times 3.5\times 10^3\\\\\Rightarrow v=\sqrt{\dfrac{2\times 3.5\times 10^3e}{m_e}}=\sqrt{\dfrac{7\times 10^3\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}\\\\\Rightarrow v=\sqrt{1.23\times 10^{15}}=3.5\times 10^7\ m/s$

To move in a straight line Force due to magnetic field must be balanced by force due to charge

$\Rightarrow F_B=F_q\\\\\Rightarrow evB=eE\\\\\Rightarrow B=\dfrac{E}{v}\\\\\Rightarrow B=\dfrac{\frac{V}{d}}{3.5\times 10^7}=\dfrac{\frac{100}{0.029}}{3.5\times 10^7}\\\\\Rightarrow B=\dfrac{3.448\times 10^3}{3.5\times 10^7}=0.985\times 10^{-4}\ T$