A particle having a charge of 10.0,mu C and m | vedclass

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(C) When a charged particle moves in a circle within a magnetic field $B$, it experiences a magnetic force $F_m = qvB$ directed towards the center of

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$10$

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(C) When a charged particle moves in a circle within a magnetic field $B$, it experiences a magnetic force $F_m = qvB$ directed towards the center of the circle.To make the particle move in a straight line with uniform velocity, the net force on it must be zero.This is achieved by applying an electric field $E$ such that the electric force $F_e = qE$ exactly balances the magnetic force $F_m$.Thus, $qE = qvB$, which simplifies to $E = vB$.The radius of the circular path is given by $r = \frac{mv}{qB}$, so the velocity is $v = \frac{qBr}{m}$.Substituting this expression for $v$ into the equation for $E$, we get $E = \left(\frac{qBr}{m}ight)B = \frac{qB^2r}{m}$.Given values: $q = 10.0\,\mu C = 10 	imes 10^{-6}\,C$, $m = 1\,\mu g = 1 	imes 10^{-9}\,kg$, $r = 10\,cm = 0.1\,m$, and $B = 0.1\,T$.Substituting these values: $E = \frac{(10 	imes 10^{-6}) 	imes (0.1)^2 	imes 0.1}{1 	imes 10^{-9}} = \frac{10^{-5} 	imes 0.01 	imes 0.1}{10^{-9}} = \frac{10^{-8}}{10^{-9}} = 10\,V/m$.

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