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 speed of the particles are $$v_A$$ and $$v_B$$ respectively, and the trajectories are as shown in the figure. Then

A
mAvA<mBvB
B
mAvA>mBvB
C
mA<mB and vA<vB
D
mA=mB and vA=vB

Solution

## The correct option is B $$m_Av_A > m_Bv_B$$When a charged particle is moving at right angle to the magnetic field, then a force acts on it which behaves as a centripetal force and moves the particle in circular path.But, $$r= \dfrac{p}{qB}$$Radius of B is less than  radius of A.$$r_B \lt r_A$$$$\therefore p_B \lt p_A$$$$\Rightarrow m_Bv_B \lt m_Av_A$$$$\Rightarrow m_Av_A > m_Bv_B$$.Physics

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