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

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 the figure. Then (A) mA vA < mB vB (B) mA vA > mB vB (C) mA < vB < and mA v < B (D) mA = vB < and mA = vB

Tardigrade Admin · Accepted Answer

Radius of the circle=(mv/Bq) or radius propto mv if B and q are same. (Radius)A > (Radius)B ∴ mA vA > mB vB

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

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

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

$m_{A} < v_{B} <$ and $m_{A} v <_{B}$

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

Radius of the circle $= \frac{m v}{Bq}$
or radius $\propto$ mv if B and q are same.
$(R a d i u s)_{A} > (R a d i u s)_{B}$
$∴ m_{A} v_{A} > m_{B} v_{B}$

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 the figure. Then

mA​vA​<mB​vB​

mA​vA​>mB​vB​

mA​<vB​< and mA​v<B​

mA​=vB​< and mA​=vB​

Radius of the circle=Bqmv​ or radius ∝ mv if B and q are same. (Radius)A​>(Radius)B​ ∴mA​vA​>mB​vB​

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

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

$m_{A} < v_{B} <$ and $m_{A} v <_{B}$

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

Radius of the circle $= \frac{m v}{Bq}$
or radius $\propto$ mv if B and q are same.
$(R a d i u s)_{A} > (R a d i u s)_{B}$
$∴ m_{A} v_{A} > m_{B} v_{B}$