Question

Two identical particles having the same mass $m$ and charges $+q$ and $-q$, separated by a distance $d$, enter in a uniform magnetic field $B$ directed perpendicular to paper inward with speeds $v_1$ and $v_2$ as shown in the figure. The particles will not collide if

• A. $d>\frac{m}{Bq}(v_1+v_2)$
• B. $d>\frac{m}{Bq}(v_1+v_2)$
• C. $d>\frac{2m}{Bq}(v_1+v_2)$
• D. $v_1=v_2$

Answer 1

The correct option is C $d>\frac{2m}{Bq}(v_1+v_2)$

The particles $+q$ and $-q$ will experience some force due to magnetic field and the direction of this force can be determined by Fleming's left hand rule. $r_1$ and $r_2$ are the radius of circular path followed by $-q$ and $+q$ respectively.
To satisfy the condition of no collision
$d>(2r_1+2r_2)$
$⟹d>2(\frac{m{v}_1}{qB}+\frac{m{v}_2}{qB})$
$⟹d>2m(\frac{v_1+v_2}{qB}$)