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## QuestionPhysicsClass 12

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 $${E}={120}{k}{V}{m}^{{-{{1}}}}{\quad\text{and}\quad}{B}={50}{m}{T}.$$ Then the beam strikes a grounded target. Find the force which the beam acts on the target if the beam current is equal to $${i}={0.8}{m}{A}$$.
Mass of protons $$={1.67}\times{10}^{{-{27}}}{k}{g}$$.

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