A straight, long wire carries a current of 20 A . A second wire carrying an equal amount of current is placed parallel to the first. If the force exerted on a 10 cm length of the second wire is 2.0 × 10^ -5 N , what is

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Quantrex Academy · Accepted Answer

Correct answer: D. 40 cm

The force per unit length between two parallel current-carrying wires is given by

\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2 \pi d}

.

The total force on a length

L

of the second wire is

F = \dfrac{\mu_0 I_1 I_2 L}{2 \pi d}

.

Given

I_1 = 20 \text{ A}

,

I_2 = 20 \text{ A}

,

L = 10 \text{ cm} = 0.1 \text{ m}

, and

F = 2.0 \times 10^{-5} \text{ N}

.

Substituting the values into the formula:

2.0 \times 10^{-5} = \dfrac{4 \pi \times 10^{-7} \times 20 \times 20 \times 0.1}{2 \pi d}

2.0 \times 10^{-5} = \dfrac{2 \times 10^{-7} \times 40}{d}

2.0 \times 10^{-5} = \dfrac{8 \times 10^{-6}}{d}

d = \dfrac{8 \times 10^{-6}}{2.0 \times 10^{-5}} = 0.4 \text{ m}

d = 40 \text{ cm}

A straight, long wire carries a current of 20 A . A second wire...

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