Question

A particle of positive charge $q$ and mass $m$ enters with velocity $V\hat{j}$ at the origin in a magnetic field $B(-\hat{k})$ which is present in the whole space. The charge makes a perfectly inelastic collision with an identical particle $($having same charge $)$ at rest but free to move at its maximum positive $y-$coordinate. After collision, the combined charge will move on trajectory $(\text{where}r=\frac{mV}{qB})$ :

• A. $y=\frac{mv}{qV}x$
• B. $(x+r{)}^2+{(y-\frac{r}{2})}^2=\frac{r^2}{4}$
• C. $(x+r{)}^2+{(y-\frac{r}{2})}^2=\frac{r^2}{8}$
• D. $(x-r{)}^2+{(y+\frac{r}{2})}^2=\frac{r^2}{4}$

Answer 1

Answer: (B)

$(x+r{)}^2+{(y-\frac{r}{2})}^2=\frac{r^2}{4}$

Since the collision is perfectly inelastic, both the particles will stick together after collision.
Total momentum after collision will be same as total momentum before collision.
Since both particles stick together, total mass will be $2m$ and total charge will be $2q$.
$r=\frac{mv}{qB}$
$r_{before}=\frac{mv}{qB}=r$ as given
$r_{after}=\frac{mv}{2qB}$
$\therefore r_{after}=\frac{r}{2}$
So, the trajectory of the combined charge:
$(x+r{)}^2+(y-\frac{r}{2}{)}^2=(\frac{r}{2}{)}^2$