Solution
Problem Analysis:
We have a parallel plate capacitor with plate separation $d$ and potential difference $V$. Electrons are emitted from the negative plate with negligible velocity. A magnetic field $B$ is applied perpendicular to the electric field. We need to find the minimum $B$ such that no current is registered (electrons fail to reach the positive plate).
Step 1: Motion Analysis
The electrons move under the influence of crossed Electric ($E = V/d$) and Magnetic fields ($B$). Starting from rest, the path described is a cycloid. The electrons accelerate towards the positive plate but are deflected by the magnetic field.
The condition for “no current” is that the electrons just graze the positive plate but do not hit it. This corresponds to the maximum reach of the cycloid trajectory being equal to the separation $d$.
Step 2: Cycloid Maximum Reach
The maximum distance an electron travels from the cathode (negative plate) in a crossed field setup starting from rest is given by:
$$ x_{max} = \frac{2mE}{eB^2} $$To stop current, this maximum distance must not exceed the plate separation $d$:
$$ d = \frac{2mE}{eB^2} $$Step 3: Solving for B
Substitute $E = V/d$ into the equation:
$$ d = \frac{2m(V/d)}{eB^2} $$ $$ d^2 = \frac{2mV}{eB^2} $$ $$ B^2 = \frac{2mV}{ed^2} $$