Question

A potential difference of 600 V is applied across the plates of a parallel plate capacitor placed in a magnetic field. The separation between the plates is 3 mm. An electron projected vertically upward, parallel to the plates, with a velocity of $2\times {10}^{6}m{s}^{-1}$ moves undeflected between the plates. Find the magnitude and direction of the magnetic field in the region between the capacitor plates. Find the magnitude and direction of the magnetic field in the region between the capacitor plates. Given the Charge of the electron $=-1.6\times {10}^{-19}C$.

Answer 1

The force on electron will be toward the left plane due to electric field and will be equal to $F_e=eE$.
For the electron to move undeflected between the plates, there should be a force (magnetic) which is equal to the electric force and opposite in direction to electric force. So the force should be directed toward the right as the electric force is toward the left. On applying Fleming's left hand rule, we get that the magnetic field should be directed perpendicular to the plane of paper and inwards. Therefore,
Force due to electric field $=$ Force due to magnetic field
$eE=evB[\because E=\frac{V}{d}]$
$B=\frac{E}{v}=\frac{V/d}{v}$
where, $V=$ potential difference between the plates, and $d=$ distance between the plates
$B=\frac{600/3\times {10}^{-3}}{2\times {10}^6}=\frac{600}{3\times {10}^{-3}\times 2\times {10}^6}\Rightarrow B=0.1T$