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

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Question

(A) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.Substituting the given values: $m = 1.67

Vedclass · Accepted Answer

$0.148 \, m$

Vedclass · Answer

(A) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.Substituting the given values: $m = 1.67 	imes 10^{-27} \, kg$,$v = 10^7 \, m/s$,$q = 1.6 	imes 10^{-19} \, C$,and $B = 1 \, T$.$R = \frac{1.67 	imes 10^{-27} 	imes 10^7}{1.6 	imes 10^{-19} 	imes 1} = \frac{1.67}{1.6} 	imes 10^{-1} \approx 1.04375 	imes 10^{-1} \, m = 0.104375 \, m$.From the geometry of the path,the particle enters at $C$ with an angle of $45^\circ$ with the boundary. The center of the circular path lies on a line perpendicular to the velocity at $C$.The distance $CD$ is the chord length of the circular arc. The angle subtended by the chord $CD$ at the center is $2 	imes 45^\circ = 90^\circ$.Thus,$CD = 2R \sin(\frac{90^\circ}{2}) = 2R \sin(45^\circ) = 2R 	imes \frac{1}{\sqrt{2}} = R\sqrt{2}$.$CD = 0.104375 	imes 1.414 \approx 0.1476 \, m \approx 0.148 \, m$.

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