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Byju's Answer

Standard XII

Physics

Circular Kinematics

Two particles...

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

m_{A} v_{A} < m_{B} v_{B}

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m_{A} v_{A} > m_{B} v_{B}

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m_{A} < m_{B}

and

v_{A} < v_{B}

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m_{A} = m_{B}

and

v_{A} = v_{B}

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Solution

The correct option is B $m_{A} v_{A} > m_{B} v_{B}$
$r = \frac{m v}{q B} \Rightarrow r \propto m v$ (q and B are constant)
$∵ r_{A} > r_{B} \Rightarrow m_{A} v_{A} > m_{B} v_{B}$

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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

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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

The correct option is B $m_{A} v_{A} > m_{B} v_{B}$
$r = \frac{m v}{q B} \Rightarrow r \propto m v$ (q and B are constant)
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