A particle of mass 1 times 10^{-27} kg and c | vedclass

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(B) Given: mass $m = 1 	imes 10^{-27} \ kg$,charge $q = 1 	imes 10^{-16} \ C$,speed $v = 1000 \ m/s$,angle $	heta = 60^{\circ}$,number of turns $N

Vedclass · Accepted Answer

$1 	imes 10^6$

Vedclass · Answer

(B) Given: mass $m = 1 	imes 10^{-27} \ kg$,charge $q = 1 	imes 10^{-16} \ C$,speed $v = 1000 \ m/s$,angle $	heta = 60^{\circ}$,number of turns $N = 5000$,current $I = 5 \ A$.The velocity components are:$v_{\parallel} = v \cos 60^{\circ} = 1000 	imes 0.5 = 500 \ m/s$ (along the axis)$v_{\perp} = v \sin 60^{\circ} = 1000 	imes \frac{\sqrt{3}}{2} = 500\sqrt{3} \ m/s$ (perpendicular to the axis)The magnetic field inside the solenoid is $B = \mu_0 n I = \mu_0 (N/L) I$.The time taken to travel the length $L$ of the solenoid is $t = \frac{L}{v_{\parallel}} = \frac{L}{500}$.The time period of one revolution in the magnetic field is $T = \frac{2\pi m}{qB}$.The number of revolutions $n'$ is given by $n' = \frac{t}{T} = \frac{L/v_{\parallel}}{2\pi m / qB} = \frac{L \cdot q \cdot B}{v_{\parallel} \cdot 2\pi m}$.Substituting $B = \frac{\mu_0 N I}{L}$:$n' = \frac{L \cdot q \cdot (\mu_0 N I / L)}{v_{\parallel} \cdot 2\pi m} = \frac{q \mu_0 N I}{v_{\parallel} \cdot 2\pi m}$.Plugging in the values:$n' = \frac{10^{-16} 	imes (4\pi 	imes 10^{-7}) 	imes 5000 	imes 5}{500 	imes 2\pi 	imes 10^{-27}}$$n' = \frac{10^{-16} 	imes 4\pi 	imes 10^{-7} 	imes 25000}{1000\pi 	imes 10^{-27}}$$n' = \frac{10^{-16} 	imes 10^{-7} 	imes 10^5}{10^{-27} 	imes 10^2} = \frac{10^{-18}}{10^{-25}} = 10^7 / 10 = 10^6$.Thus,the number of revolutions is $1 	imes 10^6$.

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