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

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Question

(A) The time period of a charged particle moving in a uniform magnetic field is given by $T = \frac{2\pi m}{qB}$.Substituting the given values: $m = 1

Vedclass · Accepted Answer

$16.3$

Vedclass · Answer

(A) The time period of a charged particle moving in a uniform magnetic field is given by $T = \frac{2\pi m}{qB}$.Substituting the given values: $m = 1.67 	imes 10^{-27} \, kg$,$q = 1.6 	imes 10^{-19} \, C$,and $B = 1 \, T$.$T = \frac{2 	imes 3.14 	imes 1.67 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes 1} \approx 6.56 	imes 10^{-8} \, s = 65.6 \, ns$.The particle enters at $45^{\circ}$ and exits at $45^{\circ}$. The angle turned by the particle inside the magnetic field is $\Delta \phi = 180^{\circ} - (45^{\circ} + 45^{\circ}) = 90^{\circ}$ (or $\frac{\pi}{2}$ radians).The time spent in the magnetic field is $t = \frac{\Delta \phi}{2\pi} 	imes T = \frac{90^{\circ}}{360^{\circ}} 	imes T = \frac{T}{4}$.$t = \frac{65.6 \, ns}{4} = 16.4 \, ns$.Comparing this with the given options,the closest value is $16.3 \, ns$.

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