Question

Two long, parallel wires are separated by a distance of 2.50 cm. The force per unit length that each wire exerts on the other is 4.00×105N/m4.00 \times 10^{-5} N/m, and the wires repel each other. The current in one wire is 0.600 A. (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions?

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Consider two long, parallel wires which are separated by a distance of 2.50 cm. It is given that the force per unit length that each wire exerts on the other is F/L=4.00×105F/L=4.00 \times 10^{-5} N/m. The two wires repel each others, and the current in one wire is I=0.600I=0.600 A.

a.

We need to find the current in the other wire. The magnitude of the force per unit length between two wires carrying currents of II and II^{\prime} is given by,

FL=μ0II2πr\begin{align}\frac{F}{L}=\frac{\mu_{0} I^{\prime} I}{2 \pi r}\end{align}

solve for II^{\prime} and substitute to get,

I=FL2πrμ0I=(4.0×105 N/m)2π(0.0250 m)(4π×107 Tm/A)(0.60 A)=8.33 A\begin{align*}I^{\prime}&=\frac{F}{L} \frac{2 \pi r}{\mu_{0} I}\\ &=\left(4.0 \times 10^{-5} \mathrm{~N/m}\right) \frac{2 \pi(0.0250 \mathrm{~m})}{(4 \pi \times 10^{-7} \mathrm{~T \cdot m/A})(0.60 \mathrm{~A})}\\ &=8.33 \mathrm{~A}\end{align*}

I=8.33 A\boxed{I^{\prime}=8.33 \mathrm{~A}}

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