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 figure, then

Question

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 figure, then (A) mAvA < mBvB (B) mAvA > mBvB (C) mA < mB and vA < vB (D) mAvA = mBvB

Tardigrade Admin · Accepted Answer

Radius R = (mv/Bq) i.e. R propto mV if B and q are fixed As radius of A > radius of B ∴ mAvA > mB vB

Q. 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 figure, then

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

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

$m_{A} < m_{B} an d v_{A} < v_{B}$

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

Radius R = $\frac{m v}{Bq} i . e . R \propto mV$ if B and $q$ are fixed
As radius of A > radius of B $∴$ $m_{A} v_{A} > m_{B} v_{B}$

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 figure, then

mA​vA​<mB​vB​

mA​vA​>mB​vB​

mA​<mB​andvA​<vB​

mA​vA​=mB​vB​