Question 7: Magnetism and Matter

Question

A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v}=\left(\frac{v_0}{\sqrt{2}} \hat{x}+\frac{v_0}{\sqrt{2}} \hat{y}\right)$. There exists a uniform magnetic field $\vec{B}=B_0 \hat{z}$ and a space varying electric field $\vec{E}=E_{\mathrm{o}} \mathrm{e}^{-\lambda x} \hat{x}$ within the region $0 \leqslant x \leqslant L$. After travelling a distance such that $x$-coordinate has changed from $x=0$ to $x=L$, the change in the kinetic energy is $\_\_\_\_$ .

Accepted Answer

The magnetic force $\vec{F}_m$ acting on a charged particle is given by $\vec{F}_m=q(\vec{v} 	imes \vec{B})$.
So, the magnetic force is always perpendicular to the velocity of the particle $\left(\overrightarrow{\mathrm{F}}_{\mathrm{m}} \perp \overrightarrow{\mathrm{v}}ight)$.
Since the force is perpendicular to the displacement at every instant, the work done by the magnetic field is always zero:
$$ \mathrm{W}_{\mathrm{m}}=\int \overrightarrow{\mathrm{F}}_{\mathrm{m}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}}=0 $$
The electric force $\overrightarrow{\mathrm{F}}_{\mathrm{e}}$ is given by $\overrightarrow{\mathrm{F}}_{\mathrm{e}}=\mathrm{q} \overrightarrow{\mathrm{E}}$.
The electric field in the region is given as $\overrightarrow{\mathrm{E}}=\mathrm{E}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{x}} \widehat{\mathrm{x}}$, the force is:
$$ \overrightarrow{\mathrm{F}}_{\mathrm{e}}=\mathrm{qE}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{x}_{\hat{\mathrm{x}}}} $$
The work done by the electric field as the particle moves from $\mathrm{x}=0$ to $\mathrm{x}=\mathrm{L}$ is :
$$ \mathrm{W}_{\mathrm{e}}=\int_0^{\mathrm{L}} \overrightarrow{\mathrm{~F}}_{\mathrm{e}} \cdot \mathrm{~d} \overrightarrow{\mathrm{x}} $$
$$\Rightarrow $$ $W_e=\int_0^L q E_o e^{-\lambda x} \cdot d x$
$\begin{gathered}\Rightarrow  W_e=e E_0 \int_0^L e^{-\lambda x} \cdot d x \ \Rightarrow  W_e=e E_0\left[\frac{e^{-\lambda x}}{-\lambda}ight]_0^L \ \Rightarrow  W_e=-\frac{q E_o}{\lambda}\left[e^{-\lambda L}-e^0ight] \ \Rightarrow W_e=-\frac{q E_o}{\lambda}\left[e^{-\lambda L}-1ight] \ \Rightarrow  W_e=\frac{q E_o}{\lambda}\left[1-e^{-\lambda L}ight]\end{gathered}$
Using work-energy theory, the total work done on particle is equal to total change in kinetic energy.
$$ \begin{gathered} \Delta K=W_{	ext {total }}=W_m+W_e \ \Delta K=0+\frac{q E_o}{\lambda}\left[1-e^{-\lambda L}ight] \ \Delta K=\frac{q E_o}{\lambda}\left[1-e^{-\lambda L}ight] \end{gathered} $$
Therefore, the change in kinetic energy is $\frac{\mathrm{qE}_0}{\lambda}\left[1-e^{-\lambda L}ight]$.
Thus, the correct option is (A).

Answer

$\frac{q E_0}{\lambda}\left[1-e^{-\lambda L}\right]$

Answer

$\left(\frac{v_0 q B_0}{2 \lambda}\right)\left[2-e^{-2 \lambda L}\right]$

Answer

$$ \frac{q E_0}{\lambda}\left[1+e^{-\lambda L}\right] $$

Answer

$$ q\left(\frac{E_0+v_0 B_0}{\lambda}\right)\left[1-e^{-\lambda L / 2}\right] $$

Magnetism and Matter

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