Question

A proton beam passes without deviation through the region of space where there exists uniform transverse mutually perpendicular electric and magnetic field with $E=120\text{kV/m}$ and $B=50\text{mT}$. The beam strikes a ground target. Find the force imparted by the beam on the target if the beam current is equal to $0.80\text{mA}$.

• A. $15.4\mu \text{N}$
• B. $19.2\mu \text{N}$
• C. $28.2\mu \text{N}$
• D. $45\mu \text{N}$

Answer 1

The correct option is B $19.2\mu \text{N}$

From Newton's $2nd$ laws of motion, the force experienced by the target is,

$F=\frac{dp}{dt}=\frac{d}{dt}\left(mv\right)....\left(1\right)$

Since the beam passes without any deflection means unaffected by $\overrightarrow{E}\&\overrightarrow{B}$, so it will travel in a straight line.

Thus, net force on the proton will be zero.

${\overrightarrow{F}}_{net}=0$

$\therefore qE=qvB\sin {90}^{\circ }=qvB$

$\therefore v=\frac{E}{B}=\text{constant}$

Now equation $\left(1\right)$ becomes

$F=v\frac{dm}{dt}=\frac{E}{B}(\frac{dm}{dq}\times \frac{dq}{dt})$

And we know that, $i=\frac{dq}{dt}$

$\therefore F=\frac{E}{B}\times i\times \left(\frac{dm}{dq}\right)......\left(i\right)$

Here,
$E=120\times {10}^3\text{N/m},B=50\times {10}^{-3}\text{T}$
$i=0.80\times {10}^{-3}\text{A}$

Mass by charge ratio,
$\frac{dm}{dq}=\frac{1.67\times {10}^{-27}\text{kg}}{1.6\times {10}^{-19}\text{C}}\approx {10}^{-8}\text{kg/C}$

Substituting in Eq. $\left(i\right)$ we get,

$F=\frac{120\times {10}^3}{50\times {10}^{-3}}\times 0.80\times {10}^{-3}\times {10}^{-8}$

$\therefore F=19.2\times {10}^{-6}\text{N}=19.2\mu \text{N}$

Therefore, option $\left(b\right)$ is the right answer.

Why this question ? Trick: The net force on proton beam must be zero, if it has to pass undeflected in a region of $\overrightarrow{E}$ and $\overrightarrow{B}$ present together.