Question

A current-carrying wire oriented north-south and laid over a compass deflects the compass 8° to the east. What is the magnitude of the magnetic field made by the current? The horizontal component of Earth's magnetic field is about 2 ✕ 10-5 tesla.

Answer 1

$2.8 \times 10^{-6}\ T$ is the magnitude of the magnetic field made by the current

Explanation:

Given data:

$\theta=8^{\circ}$

Magnetic field of earth, $B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}$

We need to find the magnetic field of wire, $B_{\text {wire }}$

The compass needle moves toward a direction of magnetic field. The current in 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 drawing, expressing the angle:

                   $\tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}$

By substituting the given values, we get

                  $B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}$

                  $B_{\text {wire}}=0.28 \times 10^{-5}=2.8 \times 10^{-6}\ T$