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(C) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2 \pi m}{qB}$.The proton completes $10$ revolutions,so th

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$0.44$

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(C) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2 \pi m}{qB}$.The proton completes $10$ revolutions,so the total time spent in the magnetic field is $t = 10T = 10 	imes \frac{2 \pi m}{qB}$.The velocity component parallel to the magnetic field is $v_{\parallel} = v \cos(60^{\circ}) = v 	imes \frac{1}{2} = \frac{v}{2}$.Since the velocity component parallel to the magnetic field remains constant,the distance $l$ covered in the direction of the field is $l = v_{\parallel} 	imes t = \frac{v}{2} 	imes 10 	imes \frac{2 \pi m}{qB} = \frac{10 \pi m v}{qB}$.Substituting the given values:$l = \frac{10 	imes 3.14 	imes 1.67 	imes 10^{-27} 	imes 4 	imes 10^{5}}{1.6 	imes 10^{-19} 	imes 0.3}$$l = \frac{20.942 	imes 10^{-21}}{0.48 	imes 10^{-19}} = \frac{20.942}{48} \approx 0.436 \, m$.Rounding to the nearest option,$l \approx 0.44 \, m$.

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