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(A) Given:Velocity of proton $v = 10^{6} \ m/s$ along the $Y$-axis.Electric field $E = 2 	imes 10^{4} \ V/m$ along the negative $X$-axis.Specific cha

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$0.5$

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(A) Given:Velocity of proton $v = 10^{6} \ m/s$ along the $Y$-axis.Electric field $E = 2 	imes 10^{4} \ V/m$ along the negative $X$-axis.Specific charge $\frac{e}{m} = 10^{8} \ C/kg$.Since the proton moves with a uniform velocity, the net force on it is zero. Thus, the magnetic force balances the electric force:$qE = qvB$$B = \frac{E}{v} = \frac{2 	imes 10^{4}}{10^{6}} = 2 	imes 10^{-2} \ T$When the electric field is switched off, the proton moves in a circular path due to the magnetic force acting as the centripetal force:$R = \frac{mv}{qB} = \frac{v}{(e/m)B}$$R = \frac{10^{6}}{10^{8} 	imes 2 	imes 10^{-2}}$$R = \frac{10^{6}}{2 	imes 10^{6}} = 0.5 \ m$Therefore, the radius of the circle is $0.5 \ m$.

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