Two long parallel wires are separated by 2.0 cm. The current in one of the wires is three times the other current. If the magnitude of the force on a 2.0-m length of one of the wires is equal to 60 μN, what is the greater of the two currents?
A) 2.0 A
B) 1.0 A
C) 3.0 A
D) 9.0 A
E) 1.5 A

Question

Quizplus · Accepted Answer

Using the formula for the force between two parallel wires:

$F = \frac{\mu_0}{4\pi} \frac{I_1 I_2 L}{d}$

where $\mu_0$ is the permeability of free space, $I_1$ and $I_2$ are the currents in the two wires, $L$ is the length of wire, and $d$ is the distance between the wires.

We know that $d = 2.0$ cm = 0.02 m and $L = 2.0$ m. We also know that $F = 60$ μN = $60 	imes 10^{-6}$ N.

Since the current in one wire is three times the current in the other wire, we can write $I_1 = 3I_2$. Plugging these values into the equation for the force and solving for $I_2$, we get:

$I_2 = \frac{F d}{\mu_0 \pi L I_1} = \frac{(60 	imes 10^{-6})(0.02)}{(4\pi 	imes 10^{-7})(2)(I_1)}$

$I_2 = \frac{0.012}{I_1}$

To maximize $I_1$, we need to minimize $I_2$. From the equation above, we see that as $I_2$ decreases, $I_1$ increases. Therefore, we want to choose the largest value of $I_1$ that makes $I_2$ smaller than 1.

Plugging in $I_2 = 1$, we get:

$1 = \frac{0.012}{I_1}$

$I_1 = 0.012$ A

Since this value of $I_1$ makes $I_2$ larger than 1, we need to increase $I_1$.

Plugging in $I_2 = 0.5$, we get:

$0.5 = \frac{0.012}{I_1}$

$I_1 = 0.024$ A

Again, this value of $I_1$ makes $I_2$ larger than 1, so we need to increase $I_1$.

Plugging in $I_2 = 0.3$, we get:

$0.3 = \frac{0.012}{I_1}$

$I_1 = 0.04$ A

This value of $I_1$ makes $I_2$ smaller than 1, so it is the largest value of $I_1$ that satisfies the condition. Therefore, the greater of the two currents is:

$I_1 = 3I_2 = 3(0.04)$ A

$I_1 = 0.12$ A

which is closest to choice (C) 3.0 A.

Two Long Parallel Wires Are Separated by 2