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(C) Since the velocity $v$ is not perpendicular to the magnetic field $B$,the path of the charged particle is a helix.The pitch of the helical path is

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a helix with a pitch $4 \ cm$

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(C) Since the velocity $v$ is not perpendicular to the magnetic field $B$,the path of the charged particle is a helix.The pitch of the helical path is given by the formula:$Pitch = (v \cos 	heta) 	imes T = (v \cos 	heta) 	imes \frac{2 \pi m}{B q}$Given:$v = 4 	imes 10^5 \ ms^{-1}$$	heta = 60^{\circ}$$B = 0.314 \ T$$m = 1.6 	imes 10^{-27} \ kg$$q = 1.6 	imes 10^{-19} \ C$ (charge of a proton)Substituting the values:$Pitch = (4 	imes 10^5 	imes \cos 60^{\circ}) 	imes \frac{2 	imes 3.14 	imes 1.6 	imes 10^{-27}}{0.314 	imes 1.6 	imes 10^{-19}}$$Pitch = (4 	imes 10^5 	imes 0.5) 	imes \frac{2 	imes 3.14 	imes 10^{-27}}{0.314 	imes 10^{-19}}$$Pitch = 2 	imes 10^5 	imes \frac{6.28 	imes 10^{-27}}{0.314 	imes 10^{-19}}$$Pitch = 2 	imes 10^5 	imes 20 	imes 10^{-8}$$Pitch = 40 	imes 10^{-3} \ m = 4 	imes 10^{-2} \ m = 4 \ cm$.

$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

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