A proton of mass 1.67 times 10^{-27} , kg and | vedclass

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(C) The velocity vector makes an angle of $30^{\circ}$ with the $Y$-axis (the direction of the magnetic field). Since the velocity has a component per

Vedclass · Accepted Answer

$A$ helix of radius $\cong 0.1 \, m$ and time period $2\pi 	imes 10^{-7} \, s$.

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(C) The velocity vector makes an angle of $30^{\circ}$ with the $Y$-axis (the direction of the magnetic field). Since the velocity has a component perpendicular to the magnetic field $(v \sin 30^{\circ})$ and a component parallel to it $(v \cos 30^{\circ})$,the path is a helix.The radius of the helix is given by $r = \frac{m v \sin 	heta}{q B}$,where $	heta = 30^{\circ}$ is the angle with the magnetic field.$r = \frac{1.67 	imes 10^{-27} 	imes 2 	imes 10^6 	imes \sin 30^{\circ}}{1.6 	imes 10^{-19} 	imes 0.104} \cong \frac{1.67 	imes 10^{-21} 	imes 0.5}{1.664 	imes 10^{-20}} \cong 0.5 \, m$ (Wait,re-calculating: $r = \frac{1.67 	imes 10^{-27} 	imes 2 	imes 10^6 	imes 0.5}{1.6 	imes 10^{-19} 	imes 0.104} = \frac{1.67 	imes 10^{-21}}{1.664 	imes 10^{-20}} \approx 0.1 \, m$).The time period is $T = \frac{2 \pi m}{q B} = \frac{2 	imes \pi 	imes 1.67 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes 0.104} \cong \frac{10.49 	imes 10^{-27}}{0.1664 	imes 10^{-18}} \cong 6.3 	imes 10^{-8} \, s \approx 2\pi 	imes 10^{-7} \, s$.

$A$ proton of mass $1.67 \times 10^{-27} \, kg$ and charge $1.6 \times 10^{-19} \, C$ is projected with a speed of $2 \times 10^6 \, m/s$ at an angle of $60^{\circ}$ to the $X$-axis. If a uniform magnetic field of $0.104 \, T$ is applied along the $Y$-axis,the path of the proton is:

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