A uniform magnetic field B is acting from sou | vedclass

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(B) The force on a charged particle moving in a magnetic field is given by $F = qvB \sin 	heta$. Since the particle moves vertically downwards and th

Vedclass · Accepted Answer

$7.4 	imes 10^{-12} \ N$

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(B) The force on a charged particle moving in a magnetic field is given by $F = qvB \sin 	heta$. Since the particle moves vertically downwards and the field is horizontal (south to north),the angle $	heta = 90^\circ$,so $\sin 90^\circ = 1$. Thus,$F = qvB$.Given kinetic energy $K = 5 \ MeV = 5 	imes 10^6 	imes 1.6 	imes 10^{-19} \ J = 8 	imes 10^{-13} \ J$.Using $K = \frac{1}{2}mv^2$,we find velocity $v = \sqrt{\frac{2K}{m}}$.Substituting $v$ into the force equation: $F = qB \sqrt{\frac{2K}{m}}$.$F = (1.6 	imes 10^{-19}) 	imes 1.5 	imes \sqrt{\frac{2 	imes 8 	imes 10^{-13}}{1.7 	imes 10^{-27}}}$.$F = 2.4 	imes 10^{-19} 	imes \sqrt{9.41 	imes 10^{14}} \approx 2.4 	imes 10^{-19} 	imes 3.067 	imes 10^7$.$F \approx 7.36 	imes 10^{-12} \ N$,which is approximately $7.4 	imes 10^{-12} \ N$.

$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be

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