Question

A non-relativistic proton beam passes without deviation through the region of space where uniform transverse mutually perpendicular electric and magnetic fields of $E=120{\text{kVm}}^{-1}$ and $B=50\text{mT}$. Then the beam strikes a grounded target. Find the force with which the beam hits on the target if the beam current is equal to $I=0.80\text{mA}$.[Take mass of proton and its charge as $m_e=1.67\times {10}^{-27}\text{kg}$ and $q=1.6\times {10}^{-19}\text{C}$ respectively.]

• A. $10\mu \text{N}$
• B. $40\mu \text{N}$
• C. $30\mu \text{N}$
• D. $20\mu \text{N}$

Answer 1

The correct option is D $20\mu \text{N}$

Given, $E=120{\text{kV m}}^{-1};B=50\text{mT};I=0.80\text{mA}$

As proton beam passes without deviation through the region of uniform magnetic field and electric fields.


$\therefore F_E=F_B$

$\Rightarrow qE=qvB\sin \theta [\because \theta ={90}^{\circ }]$

$\Rightarrow v=\frac{E}{B}=\frac{120\times {10}^3}{50\times {10}^{-3}}=2.4\times {10}^6{\text{ms}}^{-1}$

Force exerted by the proton beam on the target is, $F=\frac{dp}{dt}$

Where, $dp=$ Change in momentum of the proton beam,

$\Rightarrow F=\frac{dp}{dt}=\frac{d\left(mv\right)}{dt}$

As, there is no change in velocity,

$\therefore F=v\frac{dm}{dt}=v\left(\frac{dm}{dq}\right)\left(\frac{dq}{dt}\right)$

$\Rightarrow F=v\times I\times \frac{dm}{dq}$

$=2.4\times {10}^6\times 0.8\times {10}^{-3}\times \frac{1.67\times {10}^{-27}}{1.6\times {10}^{-19}}$

$\approx 2\times {10}^{-5}\text{N}\approx 20\mu \text{N}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.