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(D) The magnetic field $B$ produced by an infinitely long straight conductor at a distance $R$ is given by:$B = \frac{\mu_{0} I}{2 \pi R}$Given: $I =

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$8$

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(D) The magnetic field $B$ produced by an infinitely long straight conductor at a distance $R$ is given by:$B = \frac{\mu_{0} I}{2 \pi R}$Given: $I = 5 \, A$,$R = 20 \, cm = 0.2 \, m$,$v = 10^{5} \, m/s$,$q = e = 1.6 	imes 10^{-19} \, C$.Substituting the values:$B = \frac{4 \pi 	imes 10^{-7} 	imes 5}{2 \pi 	imes 0.2} = \frac{2 	imes 10^{-7} 	imes 5}{0.2} = 5 	imes 10^{-6} \, T$.The magnetic force $F$ on a moving charge is given by $F = qvB \sin 	heta$. Since the electron moves parallel to the conductor,the angle $	heta$ between the velocity vector and the magnetic field (which is perpendicular to the plane of the conductor and the electron's path) is $90^{\circ}$.$F = qvB \sin 90^{\circ} = qvB$$F = (1.6 	imes 10^{-19} \, C) 	imes (10^{5} \, m/s) 	imes (5 	imes 10^{-6} \, T)$$F = 8 	imes 10^{-20} \, N$.Thus,the magnitude of the force is $8 	imes 10^{-20} \, N$.

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