A long vertical wire carries an unknown current. Coaxial with the wire is a long, thin, cylindrical conducting surface that carries a current of$\mathbf{30}\mathbf{ }\mathbf{\text{mA}}$upward. The cylindrical surface has a radius of$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{mm}}$. If the magnitude of the magnetic field at a point$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{mm}}$from the wire is$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu T}$, (a) What are the size and(b) What is the direction of the current in the wire?

1. Current$i=30\text{mA}$
2. Radius of wire$R=3 \text{mm}$
3. Magnitude of magnetic field$B=1\mathrm{\mu T}$
4. Distance$r=5 \text{mm}$

An equation known as the Biot-Savart Law describes the magnetic field produced by a steady electric current. It connects the electric current's strength, direction, length, and proximity to the magnetic field.

We use the Biot-Savart law to find the magnetic field due to a vertical wire. Using this, we can find the current flowing through the wire and the direction of the current.

Formula:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$

The magnetic field due to a long straight wire at perpendicular distance $r$is given by

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

Where $r=5\text{mm} \text{or} 5\times {10}^{-3}\text{m}$-perpendicular distance from the wire and ${\mu }_0=1.67\times {10}^{-6}\text{T.m/A}$is the permeability of free space.

Solving for current of the wire we get

role="math" localid="1663010280645" $i=\frac{2\pi rB}{{\mu }_0}$

$i=\frac{2\pi \times 5\times {10}^{-3}{\text{m×10}}^{-6}\text{T}}{4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{TA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}}$

$i=0.025\text{A} \text{or} 25\text{mA}$

Hence, the thin wire must carry the current of$i=0.025\text{A} \text{or} 25\text{mA}$.

The direction of the current is downward and opposite to $30\mathrm{mA}$. current carried by the thin conducting surface of the wire.