A particle having a charge of 10,mu C and 1,m | vedclass

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(C) Given:Charge $q = 10\,\mu C = 10 	imes 10^{-6}\, C$Mass $m = 1\,\mu g = 1 	imes 10^{-9}\, kg$Radius $r = 10\, cm = 0.1\, m$Magnetic field $B = 0

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$10$

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(C) Given:Charge $q = 10\,\mu C = 10 	imes 10^{-6}\, C$Mass $m = 1\,\mu g = 1 	imes 10^{-9}\, kg$Radius $r = 10\, cm = 0.1\, m$Magnetic field $B = 0.1\, T$For the particle to move tangentially with a constant speed,the net force acting on it must be zero. This means the electric force must balance the magnetic force:$|\overrightarrow{F}_{m}| = |\overrightarrow{F}_{e}|$$qvB = qE \Rightarrow E = vB$From the formula for the radius of a circular path in a magnetic field:$r = \frac{mv}{qB} \Rightarrow v = \frac{qBr}{m}$Substituting the value of $v$ into the expression for $E$:$E = \left(\frac{qBr}{m}ight) B = \frac{qB^{2}r}{m}$Substituting the given values:$E = \frac{(10 	imes 10^{-6}) 	imes (0.1)^2 	imes 0.1}{1 	imes 10^{-9}}$$E = \frac{10^{-5} 	imes 0.01 	imes 0.1}{10^{-9}} = \frac{10^{-8}}{10^{-9}} = 10\, V/m$

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