Question

determine the magnitude of the force between two parallel wires, one very long and another 42.1 m - long separated by 0.0502 m, both carrying 6.98 a in the same direction.
o 4.09×10⁻³ n
o 1.17×10⁻³ n
o 1.94×10⁻⁴ n
o 8.17×10⁻³ n
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Explanation:

Step1: Recall force formula

The force between two parallel current - carrying wires is given by $F = \frac{\mu_0I_1I_2L}{2\pi r}$, where $\mu_0=4\pi\times 10^{- 7}\ T\cdot m/A$, $I_1 = I_2=6.98\ A$, $L = 42.1\ m$, and $r = 0.0502\ m$.

Step2: Substitute values

Substitute the values into the formula:

$$
LATEXBLOCK0
$$

First, calculate $6.98^{2}=48.7204$. Then $48.7204\times42.1 = 2051.12884$.

$$
LATEXBLOCK1
$$

Answer:

$8.17\times 10^{-3}\ N$