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(D) The force on a charged particle in a magnetic field is given by $F = qvB$. Since $F = ma$,we have $ma = qvB$,which implies $a = \frac{qvB}{m}$.Giv

Vedclass · Accepted Answer

$0.71$

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(D) The force on a charged particle in a magnetic field is given by $F = qvB$. Since $F = ma$,we have $ma = qvB$,which implies $a = \frac{qvB}{m}$.Given kinetic energy $K = \frac{1}{2}mv^2$,we find the velocity $v = \sqrt{\frac{2K}{m}}$.Substituting $v$ into the acceleration formula: $a = \frac{qB}{m} \sqrt{\frac{2K}{m}} = \frac{qB \sqrt{2K}}{m^{3/2}}$.Rearranging for $B$: $B = \frac{ma}{qv} = \frac{ma}{q \sqrt{2K/m}} = \frac{m^{3/2} a}{q \sqrt{2K}}$.Given values: $m = 1.6 	imes 10^{-27} \; kg$,$a = 10^{12} \; m/s^2$,$q = 1.6 	imes 10^{-19} \; C$,$K = 1 \; MeV = 10^6 	imes 1.6 	imes 10^{-19} \; J = 1.6 	imes 10^{-13} \; J$.$B = \frac{(1.6 	imes 10^{-27})^{3/2} 	imes 10^{12}}{1.6 	imes 10^{-19} 	imes \sqrt{2 	imes 1.6 	imes 10^{-13}}} = \frac{(1.6 	imes 10^{-27}) 	imes \sqrt{1.6 	imes 10^{-27}} 	imes 10^{12}}{1.6 	imes 10^{-19} 	imes \sqrt{3.2 	imes 10^{-13}}} \approx 0.71 	imes 10^{-3} \; T = 0.71 \; mT$.

$A$ proton with kinetic energy of $1 \; MeV$ moves from south to north. It experiences an acceleration of $10^{12} \; m/s^2$ due to an applied magnetic field (directed from west to east). The value of the magnetic field is: ....... $mT$ (Rest mass of proton is $1.6 \times 10^{-27} \; kg$)

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