JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 13)
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 :
1
2.4
1.4
2
Explanation
$${{dF} \over {dl}} = 2 \times {10^{ - 6}}$$ N/m $$ = {{{\mu _0}{i_1}{i_2}} \over {2\pi d}}$$
$$2 \times {10^{ - 6}} = {{2 \times {{10}^{ - 7}} \times {x^2}} \over {0.2}}$$
$$x = \sqrt 2 \simeq 1.4$$
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