Question

In figure given, an electron get accelerated from rest through potential difference $V_{1}=500 \,V$. It enters the gap between two parallel plates having separation $d=20.0\, mm$ and potential difference $V_{2}=100\,V$. The upper plate is at the higher potential. Neglecting fringing and assuming that the electron's velocity vector is perpendicular to the electric field vector between the plates, find the magnitude of uniform magnetic field (in $\times 10^{-3} T$ ) which allows the electron to travel in a straight line in the gap? (Mass of electron $m_{e}=9.0 \times 10^{-31} kg$ )

Answer 1

Straight-line motion will result from zero net force acting on the system; we ignore gravity. Due to electric field the electron experiences force in upward direction, hence due to magnetic field it will experience force in downwards direction.
Thus, $\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})=0$.
Note that $\vec{v} \perp \vec{B}$
so $|\vec{v} \times \vec{B}|=v B$.
Hence, $\vec{E}=-\vec{v} \times \vec{B}$
$ \Rightarrow|E|=|v B|$
The electric field between the plates,
$E=\frac{V_{2}}{d}=\frac{100}{20 \times 10^{-3}}=5.0 \times 10^{4} V / m$
The electron gets accelerated through potential difference $V_{1}=500\, V$.
Hence its kinetic energy, $K=e V_{1}=\frac{1}{2} m_{e} v^{2}$
The speed of the electron,
$v=\sqrt{\frac{2 e V_{1}}{m_{e}}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 500}{9.0 \times 10^{-31}}}=\frac{4}{3} \times 10^{7} m / s$
Hence magnetic field, $B=\frac{E}{v}$
$=\frac{5.0 \times 10^{4}}{4 / 3 \times 10^{7}}=3.75 \times 10^{-3} T$