Question

A non-relativistic proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic field with $E = 120kV{m^{ - 1}}$ and $B = 50mT$.Then the beam strikes the ground target. Find the force imparted by the beam on the target if the beam current is equal to $i = 0.8 \times {10^{ - 3}}A.$
(Mass of protons = $1.67 \times {10^{ - 27}}kg$)

Answer 1

Hint: Electric field is caused by stationary charges and magnetic fields are caused by moving electric charges. Stationary charges cause electric fields. Since a non-relativistic proton beam passes without deviation through a region of space, we can say that force due to magnetic field equals force due to electric field. Then we must calculate the number of photons in the beam. Determine the force imparted by the beam by applying the given data.

Formula used:
$v = \dfrac{E}{B}\left( {m{s^{ - 1}}} \right)$
$i = ne\left( A \right)$
$F = nmv\left( N \right)$
Where $v$ is the velocity of the beam, $E$ is the electric field, $B$ is the magnetic field, $i$ is the current, $n$ is the number of photons, $m$ is the mass of photons.

Complete step by step answer:
A magnetic field induces electric charges movement producing electric current.
The condition for a non-relativistic proton beam passes without deviation through a region of space is ${F_e} = {F_B}$
$eE = eBv$
Since $\left( {{F_e} = eE,{F_B} = eBv} \right)$
We will get,
$v = \dfrac{E}{B}$
$ \Rightarrow \dfrac{{120 \times {{10}^3}}}{{50 \times {{10}^{ - 3}}}}$
On simplification we get.
$ \Rightarrow 2.4 \times {10^{{6^{}}}}m/s$
Here the number of protons striking per second be \[n\], then we can write it as,
$n = \dfrac{{0.8 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 19}}}}$
$n = 15 \times {10^{{{15}^{}}}}{s^{ - 1}}$
Where charge of an electron e is $1.6 \times {10^{ - 19}}coulomb$.
Force imparted by the beam = rate of change in momentum
$F = nmv$
Putting the values we get,
$F = 5 \times {10^{{{15}^{}}}} \times 1.67 \times {10^{ - 27}} \times 2.4 \times {10^6}$
On doing some simplification we get, $F = 2 \times {10^{ - 5}}N.$

Note: Magnetic field and Electric field are always perpendicular to each other and perpendicular to the direction of propagation. Electromagnetic waves contain both electric and magnetic field waves. The electromagnetic waves oscillate in perpendicular planes with respect to each other and are in phase. Force is the rate of change of momentum.