question_answer

Figure shows a circular region of radius \[R=\sqrt{3}m\]which has a uniform magnetic field \[B=0.2\text{ }T\]directed into the plane of the figure. A particle having mass\[m=2g\], speed \[v=0.3\text{ }m/s\]and charge \[q=1\text{ }mC\]is projected along the radius of the circular region as shown in figure. Calculate the angular deviation produced in the path of the particle as it comes out of the magnetic field. Neglect any other force apart from the magnetic force.

A) \[30{}^\circ \]                   

B) \[60{}^\circ \]      

C) \[75{}^\circ \]                   

D) \[90{}^\circ \]

Correct Answer: B

Solution :

[b] Radius of circular path of the particle is \[r=\frac{mv}{qB}=\frac{2\times {{10}^{-3}}\times 0.3}{{{10}^{-3}}\times 0.2}=3m\] C is centre of the circular path.             Particle enters the field at A and leaves at B.             \[\tan \theta =\frac{R}{r}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\] Deviation \[=2\theta =60{}^\circ .\] [A careful observation tells that direction of velocity at B is along the radius of the given circle].