Question

A charged particle $\left(q\right)$ is projected in a uniform magnetic field perpendicular to it with intial velocity $\left(v_0\right)$ along x-axis from origin, as shown. A tangential force $\overrightarrow{F_1}=-\alpha \overrightarrow{v}$ always acts along its path and hence, finally the particle stops. Then

• A. Ratio of displacement to distance travelled by particle before it finally stops is $\frac{\alpha }{\sqrt{q^2{B}^2+{\alpha }^2}}$.
• B. Displacement of the particle before it finally stops is $\frac{m{V}_0}{\sqrt{q^2{B}^2+{\alpha }^2}}.$
• C. Displacement of the particle before it finally stops is $\frac{V_0}{\sqrt{4q^2{B}^2+{\alpha }^2}}.$
• D. Ratio of displacement to distance travelled by particle before it finally stops is $\frac{2m^2}{\alpha \sqrt{(qB{)}^2+2{\alpha }^2}}$

Answer 1

Answer: (A), (B)

The correct options are
A Ratio of displacement to distance travelled by particle before it finally stops is $\frac{\alpha }{\sqrt{q^2{B}^2+{\alpha }^2}}$.
B Displacement of the particle before it finally stops is $\frac{m{V}_0}{\sqrt{q^2{B}^2+{\alpha }^2}}.$


$|\overrightarrow{a_c}|=\frac{qvB}{m}$
$|\overrightarrow{a_t}|=\frac{\alpha V}{m}$
Where $\overrightarrow{a_t}$ is the rate of change of speed.

$\therefore \frac{\alpha V}{m}=\frac{vdv}{dD}$, where v is speed and D is distance covered.

By integration we get, $D=\frac{m{V}_0}{\alpha }$

For displacement,
$|\overrightarrow{a_{net}}|=\sqrt{|\overrightarrow{a_c}{|}^2+|\overrightarrow{a_t}{|}^2}$
$a_{net}=\frac{dv}{dt}=\frac{vdv}{dS}$, whereby S is displacement.
$\frac{v\sqrt{q^2{B}^2+{\alpha }^2}}{m}=\frac{vdv}{dS}$

By integration we get, $S=\frac{m{V}_0}{\sqrt{q^2{B}^2+{\alpha }^2}}$

$\therefore$ ratio of displacement and distance is $\frac{\alpha }{\sqrt{q^2{B}^2+{\alpha }^2}}$.