MCQ
Two straight parallel wires, both carrying $10$ $ampere$ in the same direction attract each other with a force of $1 \times {10^{ - 3}}\,N$. If both currents are doubled, the force of attraction will be
  • A
    $1 \times {10^{ - 3}}\,N$
  • B
    $2 \times {10^{ - 3}}\,N$
  • $4 \times {10^{ - 3}}\,N$
  • D
    $0.25 \times {10^{ - 3}}\,N$

Answer

Correct option: C.
$4 \times {10^{ - 3}}\,N$
c
(c) $F = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{i_1}{i_2}}}{a} = {10^{ - 3}}\,N$
When current in both the wires is doubled, then
$F' = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\left( {2{i_1} \times 2{i_2}} \right)}}{a} = 4 \times {10^{ - 3}}\,N$

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