A straight conductor carries a current of $5A$. An electron travelling with a speed of $5 \times {10^6}\,m{s^{ - 1}}$ parallel to the wire at a distance of $0.1\,m$ from the conductor, experiences a force of
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(c) Magnetic field produced by wire is perpendicular to the motion of electron and it is given by
$B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{a} = {10^{ - 7}} \times \frac{{2 \times 5}}{{0.1}} = {10^{ - 5}}\,Wb/{m^2}$
Hence force on electron
$F = qvB = (1.6 \times {10^{ - 19}}) \times 5 \times {10^6} \times {10^{ - 5}} = 8 \times {10^{ - 18}}\,N$
$B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{a} = {10^{ - 7}} \times \frac{{2 \times 5}}{{0.1}} = {10^{ - 5}}\,Wb/{m^2}$
Hence force on electron
$F = qvB = (1.6 \times {10^{ - 19}}) \times 5 \times {10^6} \times {10^{ - 5}} = 8 \times {10^{ - 18}}\,N$
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