A particle of charge 1.6 mu C and mass 16 m | vedclass

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(A) Given:Charge $q = 1.6 \ \mu C = 1.6 	imes 10^{-6} \ C$Mass $m = 16 \ \mu g = 16 	imes 10^{-9} \ kg$Magnetic field $B = 6.28 \ T$Since the partic

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$0.01$

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(A) Given:Charge $q = 1.6 \ \mu C = 1.6 	imes 10^{-6} \ C$Mass $m = 16 \ \mu g = 16 	imes 10^{-9} \ kg$Magnetic field $B = 6.28 \ T$Since the particle is fired perpendicular to the magnetic field,it will perform uniform circular motion.The time period $T$ for one complete revolution is given by the formula:$T = \frac{2 \pi m}{qB}$Substituting the values:$T = \frac{2 	imes 3.14 	imes 16 	imes 10^{-9}}{1.6 	imes 10^{-6} 	imes 6.28}$$T = \frac{6.28 	imes 16 	imes 10^{-9}}{1.6 	imes 10^{-6} 	imes 6.28}$$T = \frac{16 	imes 10^{-9}}{1.6 	imes 10^{-6}}$$T = 10 	imes 10^{-3} \ s = 0.01 \ s$Thus,the time required to return to the original location is $0.01 \ s$.

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

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