Problem

Two parallel wires, 10 cm apart, carry currents of $8 \, \mathrm{A}$ each in opposite directions. What is the magnetic field halfway between them?

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Data

• Current in the first wire: $I_1 = 8 \, \mathrm{A}$,
• Current in the second wire: $I_2 = 8 \, \mathrm{A}$,
• Distance halfway between the wires: $r = 5 \, \mathrm{cm} = 5 \times 10^{-2} \, \mathrm{m}$,
• Permeability of free space: $\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{Wb \cdot A^{-1} \cdot m^{-1}}$.

To find:

• Net magnetic field halfway: $B_{\text{net}} = ?$.

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Prerequisite Concepts

The magnetic field generated by a long straight current-carrying wire at a distance $r$ is given by:

$$B = \frac{\mu_0}{2 \pi} \cdot \frac{I}{r}$$

Since the currents in the wires are in opposite directions, the magnetic fields generated by each wire will add up midway between them:

$$B_{\text{net}} = B_1 + B_2 = \frac{\mu_0}{2 \pi} \cdot \frac{I_1}{r} + \frac{\mu_0}{2 \pi} \cdot \frac{I_2}{r}$$

Simplify:

$$B_{\text{net}} = \frac{\mu_0}{2 \pi r} \cdot (I_1 + I_2)$$

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Solution

1. Substitute the known values into the formula:

$$B_{\text{net}} = \frac{4 \pi \times 10^{-7}}{2 \pi \cdot 5 \times 10^{-2}} \cdot (8 + 8)$$

2. Simplify the denominator:

$$2 \pi \cdot 5 \times 10^{-2} = \pi \cdot 10^{-1}.$$

3. Simplify the numerator:

$$\frac{4 \pi \times 10^{-7}}{\pi \cdot 10^{-1}} = \frac{4 \times 10^{-7}}{10^{-1}} = 4 \times 10^{-6}.$$

4. Multiply by the sum of currents:

$$B_{\text{net}} = 4 \times 10^{-6} \cdot (8 + 8) = 4 \times 10^{-6} \cdot 16 = 6.4 \times 10^{-5} \, \mathrm{T}.$$

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Answer

The net magnetic field halfway between the wires is:

$$B_{\text{net}} = 6.4 \times 10^{-5} \, \mathrm{T}.$$