An electron has mass 9 times 10^{-31} kg and | vedclass

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(B) When a charged particle moves in a magnetic field perpendicular to its velocity,it follows a circular path.The radius $r$ of the circular path is

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$5.6 	imes 10^{-5} \ T$

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(B) When a charged particle moves in a magnetic field perpendicular to its velocity,it follows a circular path.The radius $r$ of the circular path is given by the formula: $r = \frac{mv}{qB}$.Rearranging the formula to solve for the magnetic field intensity $B$,we get: $B = \frac{mv}{qr}$.Given values:Mass of electron $(m)$ = $9 	imes 10^{-31} \ kg$Velocity $(v)$ = $10^6 \ m/s$Charge $(q)$ = $1.6 	imes 10^{-19} \ C$Radius $(r)$ = $0.10 \ m$Substituting these values into the formula:$B = \frac{9 	imes 10^{-31} 	imes 10^6}{1.6 	imes 10^{-19} 	imes 0.1}$$B = \frac{9 	imes 10^{-25}}{0.16 	imes 10^{-19}}$$B = \frac{9}{0.16} 	imes 10^{-6} \ T$$B = 56.25 	imes 10^{-6} \ T = 5.625 	imes 10^{-5} \ T$.Rounding to the nearest option,we get $B = 5.6 	imes 10^{-5} \ T$.

An electron has mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is moving with a velocity of $10^6 \ m/s$. It enters a region where a magnetic field exists. If it describes a circle of radius $0.10 \ m$,the intensity of the magnetic field must be:

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