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

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(A) The time period of a charged particle in a magnetic field is given by $T = \frac{2\pi m}{qB}$.Substituting the values: $T = \frac{2 	imes 3.14

Vedclass · Accepted Answer

$16$

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(A) The time period of a charged particle in a magnetic field is given by $T = \frac{2\pi m}{qB}$.Substituting the values: $T = \frac{2 	imes 3.14 	imes 1.67 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes 1} \approx 65.6 	imes 10^{-9} \, s = 65.6 \, ns$.The particle enters at $45^{\circ}$ and leaves at $45^{\circ}$ due to symmetry. The angle subtended by the arc at the center of the circular path is $\phi = 180^{\circ} - (45^{\circ} + 45^{\circ}) = 90^{\circ}$ (or $\frac{\pi}{2}$ radians).The time spent in the magnetic field is $t = \frac{\phi}{2\pi} 	imes T = \frac{90^{\circ}}{360^{\circ}} 	imes T = \frac{T}{4}$.$t = \frac{65.6}{4} \, ns = 16.4 \, ns$.Rounding to the nearest integer,the time is $16 \, ns$.

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