Two particles A and B of masses $m$ and $2 m$ have charges $q$ and $2 q$ , respectively. They are moving with velocities $v_{1}$ and $v_{2}$ , respectively, in the same direction, enters the same magnetic field, $B$ acting normally to their direction of motion. If the two forces $F_{A}$ and $F_{B}$ acting on them are in the ratio of $1 : 2$ , find the ratio of their velocities.

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Solution

Concept used: Lorentz force
The electromagnetic force on a charged particle is called as Lorentz force. When a charged particle of mass m, moves perpendicularly in an uniform magnetic field (B)
The force on the charge, F=q(v×B). Provided q is charge on the particle and v is its velocity.
Direction of the particle changes but velocity remains constant as magnetic force is always perpendicular to the velocity.

Step 1: Given data:
Charge of A is $q$ and the charge of B is $2 q$
Charges are moving with velocities $v_{1}$ and $v_{2}$ , enters the same magnetic field, $B$ .
The ratio of forces $F_{A} : F_{B} = 1 : 2$
Step 2: Formula used:
$F o r c e o n a c h a r g e p a r t i c l e m o v i n g p e r p e n d i c u l a r t o t h e m a g n e t i c f i e l d F = q v B$
Step 3: Calculating the ratio of velocities:
$\frac{F_{A}}{F_{B}} = \frac{q_{A} v_{A} B}{q_{B} v_{B} B} = \frac{1}{2} \frac{q x v_{1}}{2 q x v_{2}} = \frac{1}{2} \frac{v_{1}}{v_{2}} = \frac{1 x 2 q}{2 x q} = \frac{1}{1}$
Therefore, the ratio of velocities of charges is $1 : 1$ .

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