A 5 text{ mg} particle carrying a charge of 5 | vedclass

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(B) The particle undergoes helical motion in a uniform magnetic field.The velocity component parallel to the magnetic field is $v_\parallel = 2 	imes

Vedclass · Accepted Answer

$6.28$

Vedclass · Answer

(B) The particle undergoes helical motion in a uniform magnetic field.The velocity component parallel to the magnetic field is $v_\parallel = 2 	imes 10^{-2} 	ext{ m/s}$ (along $\hat{k}$).The time period of one revolution is $T = \frac{2\pi m}{qB}$.Given: $m = 5 	ext{ mg} = 5 	imes 10^{-6} 	ext{ kg}$, $q = 5 	imes 10^{-6} 	ext{ C}$, $B = 0.1 	ext{ T}$.$T = \frac{2 	imes \pi 	imes 5 	imes 10^{-6}}{5 	imes 10^{-6} 	imes 0.1} = \frac{2\pi}{0.1} = 20\pi 	ext{ s}$.The pitch of the helix is $p = v_\parallel 	imes T = (2 	imes 10^{-2}) 	imes (20\pi) = 0.4\pi 	ext{ m}$.For $5$ revolutions, the total distance $\alpha$ along the $\hat{k}$ direction is $\alpha = 5 	imes p = 5 	imes 0.4\pi = 2\pi 	ext{ m}$.Using $\pi \approx 3.14$, $\alpha = 2 	imes 3.14 = 6.28 	ext{ m}$.

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