Question

A non-relativistic proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with $E=120kV{m}^{-1}$ and $B=50mT$ .Then the beam strikes a grounded target. Find the force imparted by the beam on the target if the beam current is equal to $i=0.8mA$.
Mass of protons $=1.67\times {10}^{-27}kg$

Answer 1

$F_e=F_m$
or $eE=eBv$
$\therefore v=\frac{E}{B}=\frac{120\times {10}^3}{50\times {10}^{-3}}=2.4\times {10}^{6}m/s$
Let n be the number of protons striking per second.Then,
$ne=0.8\times {10}^{-3}$
or $n=\frac{0.8\times {10}^{-3}}{1.6\times {10}^{-19}}=5\times {10}^{15}/s$
Force imparted =rate of change of momentum
=$nmv$
= $5\times {10}^{15}\times 1.67\times {10}^{-27}\times 2.4\times {10}^6$
=$2.0\times 10-5N$