Chapter 31, Problem 11SP

Compute the magnitude of the magnetic field in air at a point 6.0 cm from a long straight wire carrying a current of 9.0 A.

To determine

The magnitude of the magnetic field due to a long straight wire carrying a current of $9.0\text{ A}$ at a point 6.0 cm from the wire.

Answer to Problem 11SP

Solution:

$30\mu \text{T}$

Explanation of Solution

Given data:

The distance from the current carrying wire is $6.0\text{ cm}$.

The current through the wire is $9.0\text{ A}$.

Formula used:

Write the expression for magnetic field due to a long straight current-carrying wire:

$B=\frac{{\mu }_{0}I}{2\pi r}$

Here, $B$ is the magnetic field, $I$ is the current through the wire, $r$ is the distance from the current carrying wire, and ${\mu }_0$ is the permeabilityof free space.

Explanation:

Recall the expression for magnetic due to long straight current-carrying wire:

$B=\frac{{\mu }_{0}I}{2\pi r}$

Substitute $4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A}$ for ${\mu }_0$, $9.0\text{ A}$ for $I$, and $6.0\text{ cm}$ for $r$

$\begin{array}{c}B=\frac{\left(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A}\right)\left(9.0\text{ A}\right)}{2\pi \left(6.0\text{ cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{ T}\cdot \text{m/A}\right)\left(9.0\text{ A}\right)}{2\pi \left(6.0\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}\\ =3.0\times {10}^{-5}\text{ T}\left(\frac{{10}^6\mu \text{T}}{1\text{ T}}\right)\\ =30\mu \text{T}\end{array}$

Conclusion:

The magnitude of the magnetic field at a point 6.0 cmfrom the wireis $30\mu \text{T}$.