You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

m_{A} < m_{B}

and

v_{A} < v_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m_{A} = m_{B}

and

v_{A} = v_{B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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)
$∵ r_{A} > r_{B} \Rightarrow m_{A} v_{A} > m_{B} v_{B}$

Suggest Corrections

Similar questions

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 the figure. Then

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 the figure. Then

Q. Assertion :Two equal unlike charge particles of mass

m_{1}

and

m_{2}

are moving in a plane with speeds

u_{1}

and

u_{2}

respectively. A uniform magnetic field is applied perpendicular to the plane. Both the particle leave tracks as shown in the Figure. Then,

m_{1} u_{1} > m_{2} u_{2}

Reason: Eventhough a uniform magnetic field does no work on the moving charge, lt may accelerate the charge.

Q. Two charged particles having charges

q_{1}

and

q_{2}

and masses

m_{1}

and

m_{2}

are projected with same velocity in a region of uniform magnetic field. They follow the trajectory as shown in figure. From this, we can conclude that-

Two bodies A and B are moving, towards a plane mirror with speeds $v_{A}$ and $v_{B}$ respectively as shown in the figure. The speed of image of A respect to the body B is

Join BYJU'S Learning Program

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

The correct option is B mAvA > mBvBr=mvqB ⇒ r ∝ mv (q and B are constant) ∵ rA > rB ⇒ mAvA > mBvB

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)
$∵ r_{A} > r_{B} \Rightarrow m_{A} v_{A} > m_{B} v_{B}$