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Standard XII

Physics

Circular Kinematics

Two identical...

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

d > \frac{m}{B q} (v_{1} + v_{2})

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d > \frac{m}{B q} (v_{1} + v_{2})

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d > \frac{2 m}{B q} (v_{1} + v_{2})

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v_{1} = v_{2}

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Solution

The correct option is C $d > \frac{2 m}{B q} (v_{1} + v_{2})$
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
$d > (2 r_{1} + 2 r_{2})$
$⟹ d > 2 (\frac{m v_{1}}{q B} + \frac{m v_{2}}{q B})$
$⟹ d > 2 m (\frac{v_{1} + v_{2}}{q B}$ )

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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 v1 and v2 as shown in the figure. The particles will not collide if

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