A particle of mass m = 1.67 times 10^{-27} , | vedclass

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Question

(C) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path. Given: $m = 1.67 	imes 10^{-27}

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path. Given: $m = 1.67 	imes 10^{-27} \, kg$,$q = 1.6 	imes 10^{-19} \, C$,$v = 10^7 \, m/s$,and $B = 1 \, T$. The radius of the circular path is given by $R = \frac{mv}{qB}$. Substituting the values: $R = \frac{(1.67 	imes 10^{-27}) 	imes (10^7)}{(1.6 	imes 10^{-19}) 	imes 1} \approx 0.104 \, m = 10.4 \, cm$. From the geometry of the circular path in the magnetic field,the angle of incidence at point $C$ is $45^\circ$ with the boundary. The angle between the velocity vector and the normal to the boundary is $90^\circ - 45^\circ = 45^\circ$. Due to the symmetry of the circular arc,the angle of emergence at point $D$ with the boundary will also be $45^\circ$. Therefore,$	heta = 45^\circ$.

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