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

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(C) The magnetic field $\vec{B}$ is along the $Y$-axis. The velocity $\vec{v}$ makes an angle of $60^\circ$ with the $X$-axis,so it makes an angle $

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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 magnetic field $\vec{B}$ is along the $Y$-axis. The velocity $\vec{v}$ makes an angle of $60^\circ$ with the $X$-axis,so it makes an angle $	heta = 90^\circ - 60^\circ = 30^\circ$ with the $Y$-axis (the direction of $\vec{B}$).Since the velocity has a component parallel to the magnetic field,the path of the proton is a helix.The radius of the helix is given by $r = \frac{mv \sin 	heta}{qB}$,where $	heta = 30^\circ$ is the angle between $\vec{v}$ and $\vec{B}$.$r = \frac{1.67 	imes 10^{-27} 	imes 2 	imes 10^6 	imes \sin 30^\circ}{1.6 	imes 10^{-19} 	imes 0.104} = \frac{1.67 	imes 10^{-21} 	imes 0.5}{1.664 	imes 10^{-20}} \approx 0.1 \, m$.The time period is $T = \frac{2\pi m}{qB} = \frac{2\pi 	imes 1.67 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes 0.104} \approx 2\pi 	imes 10^{-7} \, s$.

$A$ 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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