Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 $\times$ 10$-$6 N, then the value of x is approximately :
Solution
<p>${{dF} \over {dl}} = 2 \times {10^{ - 6}}$ N/m $= {{{\mu _0}{i_1}{i_2}} \over {2\pi d}}$</p>
<p>$2 \times {10^{ - 6}} = {{2 \times {{10}^{ - 7}} \times {x^2}} \over {0.2}}$</p>
<p>$x = \sqrt 2 \simeq 1.4$</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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