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

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(A) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path.The radius $r$ of the circular path is given by

Vedclass · Accepted Answer

$5.625 	imes 10^{-5} \ T$

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(A) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path.The radius $r$ of the circular path is given by the formula: $r = \frac{mv}{qB}$.Rearranging for the magnetic field intensity $B$,we get: $B = \frac{mv}{qr}$.Given values:Mass $m = 9 	imes 10^{-31} \ kg$Velocity $v = 10^6 \ ms^{-1}$Charge $q = 1.6 	imes 10^{-19} \ C$Radius $r = 10 \ cm = 0.1 \ m$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$.Thus,the correct option is $A$.

An electron of mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ moving with a velocity of $10^6 \ ms^{-1}$ enters a magnetic field normally and describes a circle of radius $10 \ cm$. The intensity of the magnetic field is:

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