A proton of energy 200, MeV enters a magnetic | vedclass

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(B) The magnetic force on a moving charge is given by $F = qvB \sin 	heta$. Since the motion is upward and the field is horizontal (south to north),t

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$1.6 	imes 10^{-10}\,N$

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(B) The magnetic force on a moving charge is given by $F = qvB \sin 	heta$. Since the motion is upward and the field is horizontal (south to north),the angle $	heta = 90^\circ$,so $F = qvB$.Given kinetic energy $K = 200\, MeV = 200 	imes 10^6 	imes 1.6 	imes 10^{-19}\,J = 3.2 	imes 10^{-11}\,J$.Using $K = \frac{1}{2}mv^2$,we find velocity $v = \sqrt{\frac{2K}{m}}$.Substituting this into the force equation: $F = qB \sqrt{\frac{2K}{m}}$.$F = (1.6 	imes 10^{-19}) 	imes 5 	imes \sqrt{\frac{2 	imes 3.2 	imes 10^{-11}}{1.67 	imes 10^{-27}}}$.$F = 8 	imes 10^{-19} 	imes \sqrt{3.83 	imes 10^{16}} \approx 8 	imes 10^{-19} 	imes 1.95 	imes 10^8 \approx 1.56 	imes 10^{-10}\,N$.Rounding to the nearest option,$F = 1.6 	imes 10^{-10}\,N$.

$A$ proton of energy $200\, MeV$ enters a magnetic field of $5\, T$. If the direction of the field is from south to north and the motion is upward,the force acting on it will be:

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