If a proton of kinetic energy 8.35 text{ MeV} | vedclass

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Question

(C) The kinetic energy of the proton is $K = 8.35 	ext{ MeV} = 8.35 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 1.336 	imes 10^{-12} 	ext{ J

Vedclass · Accepted Answer

$64 	imes 10^{-12} 	ext{ N}$

Vedclass · Answer

(C) The kinetic energy of the proton is $K = 8.35 	ext{ MeV} = 8.35 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 1.336 	imes 10^{-12} 	ext{ J}$.Using the formula for kinetic energy $K = \frac{1}{2}mv^2$,we find the velocity $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 	imes 1.336 	imes 10^{-12}}{1.67 	imes 10^{-27}}} = \sqrt{1.6 	imes 10^{15}} = \sqrt{16 	imes 10^{14}} = 4 	imes 10^7 	ext{ m/s}$.The magnetic force acting on a charged particle moving in a magnetic field is given by $F = qvB \sin(	heta)$.Since the proton enters at right angles,$	heta = 90^\circ$,so $\sin(90^\circ) = 1$.Substituting the values: $F = (1.6 	imes 10^{-19} 	ext{ C}) 	imes (4 	imes 10^7 	ext{ m/s}) 	imes (10 	ext{ T}) = 6.4 	imes 10^{-11} 	ext{ N} = 64 	imes 10^{-12} 	ext{ N}$.Thus,the correct option is $C$.

If a proton of kinetic energy $8.35 \text{ MeV}$ enters a uniform magnetic field of $10 \text{ T}$ at right angles to the direction of the field,then the force acting on the proton is (Mass of proton $= 1.67 \times 10^{-27} \text{ kg}$ and charge of proton $= 1.6 \times 10^{-19} \text{ C}$)

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