91Ó°ÊÓ

1. Current$i=30\text{mA}$
2. Radius of wire$R=3 \text{mm}$
3. Magnitude of magnetic field$B=1\mathrm{μ°Õ}$
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:

$\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{π}\mathbf{r}}$

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

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

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

Solving for current of the wire we get

role="math" localid="1663010280645" $i=\frac{2\mathrm{π}rB}{{\mathrm{μ}}_0}$

$i=\frac{2\mathrm{Ï€}×5×{10}^{-3}{\text{³¾Ã—10}}^{-6}\text{T}}{4\mathrm{Ï€}×{10}^{-7}{\displaystyle \raisebox{1ex}{$\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.