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

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$A$ helix of radius $0.1 \ m$ 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 $\vec{B}$). Since the velocity has a component parallel to the magnetic field,the path of the proton will be a helix.The radius of the helical path is given by $r = \frac{mv_{\perp}}{qB} = \frac{mv \sin(30^{\circ})}{qB}$.Substituting the values: $m = 1.67 	imes 10^{-27} \ kg$,$v = 2 	imes 10^6 \ m/s$,$\sin(30^{\circ}) = 0.5$,$q = 1.6 	imes 10^{-19} \ C$,and $B = 0.104 \ T$.$r = \frac{1.67 	imes 10^{-27} 	imes 2 	imes 10^6 	imes 0.5}{1.6 	imes 10^{-19} 	imes 0.104} \approx 0.1 \ m$.The time period of the helical motion is $T = \frac{2\pi m}{qB}$.$T = \frac{2 	imes \pi 	imes 1.67 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes 0.104} \approx 2\pi 	imes 10^{-7} \ s$.Thus,the path is a helix of radius $0.1 \ m$ and time period $2\pi 	imes 10^{-7} \ s$.

$A$ proton 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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