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(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $a$ is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{a}$

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$8 	imes 10^{-18} \, N$

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(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $a$ is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{a}$.Substituting the given values: $i = 5 \, A$,$a = 0.1 \, m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.$B = 10^{-7} 	imes \frac{2 	imes 5}{0.1} = 10^{-5} \, T$.The force $F$ on a moving charge $q$ in a magnetic field $B$ is given by $F = qvB \sin 	heta$. Since the electron moves parallel to the wire,the velocity vector is perpendicular to the magnetic field lines,so $	heta = 90^\circ$ and $\sin 90^\circ = 1$.$F = (1.6 	imes 10^{-19} \, C) 	imes (5 	imes 10^6 \, m/s) 	imes (10^{-5} \, T) = 8 	imes 10^{-18} \, N$.

$A$ straight conductor carries a current of $5 \, A$. An electron travelling with a speed of $5 \times 10^6 \, m/s$ parallel to the wire at a distance of $0.1 \, m$ from the conductor,experiences a force of:

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