Question

A positively charged particle of charge $q$ , mass $m$ enters into a uniform magnetic field with velocity $v$ as shown in figure. There is no magnetic field to the left of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{PQ}$.
Then impulse of magnetic force will be:

• A. zero
• B. $-2mv\cos \theta \hat{j}$
• C. $-2mv\sin \theta \hat{i}$
• D. $mv\sin \theta \hat{j}$

Answer 1

The correct option is C $-2mv\sin \theta \hat{i}$

From impulse momentum theorem,

$\overrightarrow{J}=\Delta \overrightarrow{P}$

Now we know that the motion of change particle in magnetic field is symmetrical.

In this case at entry position the velocity is making angle $\theta$ w.r.t boundary hence at emergence as well it will make angel $\theta$ w.r.t the boundary $\text{PQ}$



The speed at <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{A}$ and $\text{B}$ will be same as shown in figure. Since $W_{mag}=0$


${\overrightarrow{v}}_i$$=+v\sin \theta \hat{i}+v\cos \theta \hat{j}$

${\overrightarrow{v}}_f$$=-v\sin \theta \hat{i}+v\cos \theta \hat{j}$

Thus impulse is,

$\overrightarrow{J}=m({\overrightarrow{v}}_f-{\overrightarrow{v}}_i)$

$\overrightarrow{J}=m(-v\sin \theta \hat{i}+v\cos \theta \hat{j})-(v\sin \theta \hat{i}+v\cos \theta \hat{j})]$

$\therefore \overrightarrow{J}=-2mv\sin \theta \hat{i}$

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

Why this question ? Tip: If we look at the deflection emergence of charge, it is clear that only $x$-component of velocity is changing. Hence impulse of magnetic force will be along x-direction only.