A current-carrying wire oriented north-south and laid over a compass deflects the compass ${\mathbf{15}}^{\mathbf{0}}$east. What are the magnitude and direction of the magnetic field made by the current? The horizontal component of Earth’s magnetic field is about ${\mathbf{\text{2×10}}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{T}$ .

The given data can be listed below as,

• The deflection angle of a compass towards east is, $\mathrm{\theta }={15}^{\mathrm{o}}$.
• The horizontal component of Earth’s magnetic field is, ${\mathrm{B}}_{\mathrm{Earth}}=2\times {10}^{-15}\mathrm{T}$.

Whenever a specific current flow in a conducting wire, then there would be an electric field generated around the wire. The relation between the magnetic field and the current flow is a direct linear one.

The relation of magnitude of the magnetic field made by the current is expressed as,

$\mathrm{tan\theta }=\frac{{\mathrm{B}}_{\mathrm{wire}}}{{\mathrm{B}}_{\mathrm{Earth}}}$

Here, ${\mathrm{B}}_{\mathrm{wire}}$is the magnitude of the magnetic field made by the current.

Substitute all the known values in the above equation.
$\begin{array}{rcl}\tan {15}^{\mathrm{o}}& =& \left(\frac{{\mathrm{B}}_{\mathrm{wire}}}{2\times {10}^{-5}\mathrm{T}}\right)\\ {\mathrm{B}}_{\mathrm{wire}}& =& \left(2\times {10}^{-5}\mathrm{T}\right)\tan {15}^{\mathrm{o}}\\ & \approx & 5.36\times {10}^{-6}\mathrm{T}\end{array}$

The compass needle moves toward the direction of the magnetic field. The current in the wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached figure.

Thus, the magnitude of the magnetic field made by the current is $5.36\times {10}^{-6}\mathrm{T}$and the current in wire makes a magnetic field in available space where the compass is on the ground.