CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

167723.png


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image