An electron having mass 9.1 times 10^{-31} k | vedclass

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(C) The radius $r$ of the circular path of a charged particle in a magnetic field is given by the formula: $r = \frac{mv}{Bq}$.Rearranging the formula

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$2.84$

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(C) The radius $r$ of the circular path of a charged particle in a magnetic field is given by the formula: $r = \frac{mv}{Bq}$.Rearranging the formula to solve for the magnetic field intensity $B$,we get: $B = \frac{mv}{qr}$.Given values are:Mass $m = 9.1 	imes 10^{-31} \ kg$Velocity $v = 10^6 \ ms^{-1}$Charge $q = 1.6 	imes 10^{-19} \ C$Radius $r = 0.2 \ m$Substituting these values into the equation:$B = \frac{9.1 	imes 10^{-31} 	imes 10^6}{1.6 	imes 10^{-19} 	imes 0.2}$$B = \frac{9.1 	imes 10^{-25}}{0.32 	imes 10^{-19}}$$B = 28.4375 	imes 10^{-6} \ T = 2.84 	imes 10^{-5} \ T$.Thus,the intensity of the magnetic field is $2.84 	imes 10^{-5} \ T$.

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