Problem

At what distance from a long straight wire carrying a current of $10 \, \mathrm{A}$ is the magnetic field equal to the Earth’s magnetic field of $5 \times 10^{-5} \, \mathrm{T}$?

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Data

• Current in the wire: $I = 10 \, \mathrm{A}$,
• Magnetic field strength: $B = 5 \times 10^{-5} \, \mathrm{T}$,
• Permeability of free space: $\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{Wb \cdot A^{-1} \cdot m^{-1}}$.

To find:

• Distance from the wire: $r = ?$.

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Prerequisite Concepts

The magnetic field around a long straight current-carrying wire is given by Ampere’s circuital law:

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

Rearranging for $r$:

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

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Solution

1. Substitute the known values:

$$r = \frac{4 \pi \times 10^{-7}}{2 \pi} \cdot \frac{10}{5 \times 10^{-5}}$$

2. Simplify step-by-step:
   • First, calculate $\frac{\mu_0}{2 \pi}$:

$$\frac{\mu_0}{2 \pi} = \frac{4 \pi \times 10^{-7}}{2 \pi} = 2 \times 10^{-7}$$

• Next, substitute into the formula:

$$r = (2 \times 10^{-7}) \cdot \frac{10}{5 \times 10^{-5}}$$

• Simplify the fraction:

$$\frac{10}{5 \times 10^{-5}} = 2 \times 10^{4}$$

• Multiply:

$$r = (2 \times 10^{-7}) \cdot (2 \times 10^{4}) = 4 \times 10^{-3} \, \mathrm{m}$$

3. Convert to meters:

$$r = 0.04 \, \mathrm{m}$$

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Answer

The distance from the wire is $0.04 \, \mathrm{m}$ or $4 \, \mathrm{cm}$.