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Question

(C) The radius of the circular path of the charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.Given the width of the magnetic field

Vedclass · Accepted Answer

$90^{\circ}$

Vedclass · Answer

(C) The radius of the circular path of the charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.Given the width of the magnetic field region is $d = \frac{3mv}{5qB}$.Let $\alpha$ be the angle of deviation. From the geometry of the circular path,the sine of the angle of deviation $\alpha$ is given by $\sin \alpha = \frac{d}{R}$.Substituting the values,$\sin \alpha = \frac{3mv/5qB}{mv/qB} = \frac{3}{5} = 0.6$.Since $\sin 37^{\circ} = 0.6$,we have $\alpha = 37^{\circ}$.From the geometry,the angle of incidence with the normal is $i = 90^{\circ} - 53^{\circ} = 37^{\circ}$.The angle of emergence with the normal is $e = 90^{\circ} - 	heta$.The angle of deviation $\alpha$ is related to the incidence and emergence angles by $\alpha = i + e$ (for a symmetric path deviation) or more generally,the total change in direction is $\alpha$. Looking at the geometry,the particle enters at $53^{\circ}$ to the boundary and exits at $	heta$ to the boundary. The total deviation is $\alpha = 37^{\circ}$.From the geometry,the angle $	heta$ is such that the total deviation $\alpha = 180^{\circ} - (53^{\circ} + 	heta) = 37^{\circ}$.Thus,$180^{\circ} - 53^{\circ} - 	heta = 37^{\circ} \Rightarrow 127^{\circ} - 	heta = 37^{\circ} \Rightarrow 	heta = 90^{\circ}$.

$A$ particle moving with velocity $v$ having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d = \frac{3mv}{5qB}$ at an angle of $53^{\circ}$ to the boundary of the magnetic field. Find the angle $\theta$ shown in the diagram.

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