Two particles A and B of masses and respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are and respectively and the trajectories are as shown in the f

Question

Two particles A and B of masses $m_{A}$ and $m_{B}$ respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_{A}$ and $v_{B}$ respectively and the trajectories are as shown in the figure. Then [2001]

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(2)

When a charged particle is moving normal to the magnetic field, then a force acts on it which behaves as a centripetal force and moves the particle in circular motion.

$∴$ $F = q v B = \frac{m v^{2}}{r}$ $\Rightarrow r = \frac{m v^{2}}{q B}$

If $q$ and $B$ are same for both, then, $r \propto m v$

$Since r_{A} > r_{B}$

$\Rightarrow m_{A} v_{A} > m_{B} v_{B}$

Solution

1 $m_{A} v_{A} < m_{B} v_{B}$

2 $m_{A} v_{A} > m_{B} v_{B}$

3 $m_{A} < m_{B}$ and $v_{A} < v_{B}$

4 $m_{A} = m_{B}$ and $v_{A} = v_{B}$

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Solution

Q. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in the figure. Then [2001]

1 mAvA<mBvB

2 mAvA>mBvB

3 mA<mB and vA<vB

4 mA=mB and vA=vB

Ans. (2) When a charged particle is moving normal to the magnetic field, then a force acts on it which behaves as a centripetal force and moves the particle in circular motion. ∴ F=qvB=mv2r⇒r=mv2qB If q and B are same for both, then, r∝mv Since rA>rB ⇒mAvA>mBvB

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1 $m_{A} v_{A} < m_{B} v_{B}$

2 $m_{A} v_{A} > m_{B} v_{B}$

3 $m_{A} < m_{B}$ and $v_{A} < v_{B}$

4 $m_{A} = m_{B}$ and $v_{A} = v_{B}$