A straight wire is aligned north-south in a region where Earth's magnetic field \(\overrightarrow{\mathbf{B}}\) is directed \(58.0^{\circ}\) above the horizontal, with the horizontal component directed due north. The wire carries a current of \(8.00 \mathrm{A}\) toward the south. The magnetic force on the wire per unit length of wire has magnitude $2.80 \times 10^{-3} \mathrm{N} / \mathrm{m} .$ (a) What is the direction of the magnetic force on the wire? (b) What is the magnitude of \(\mathbf{B} ?\)

We are given the following values: - The current \(I = 8.00\,\text{A}\) is flowing toward the south - Earth's magnetic field \(\overrightarrow{\mathbf{B}}\): \(58.0^{\circ}\) above the horizontal, with the horizontal component directed due north - The magnetic force on the wire per unit length has a magnitude \(2.80 \times 10^{-3}\, \text{N/m}\)

To find the direction of the magnetic force on the wire, we need to use the right-hand rule, as we are given that the current is flowing towards the south: 1. Point the thumb of your right hand in the direction of the wire's current (towards the south) 2. Point the other fingers of your right hand in the direction of the magnetic field The palm of your hand now points in the direction of the magnetic force. In our case, this is towards the ground. Therefore, the magnetic force on the wire is directed downward.

Let's now find the magnitude of the magnetic field \(\overrightarrow{\mathbf{B}}\) by using the given relationship between the force, the current and the magnetic field. Since \(\overrightarrow{F} = I\overrightarrow{L} \times \overrightarrow{B}\), we know the force per unit length on the wire is: \(F = |I\overrightarrow{L} \times \overrightarrow{B}|\) The force is given as a magnitude per unit length, \(F = 2.80 \times 10^{-3}\, \text{N/m}\). We can rewrite the equation as: \(|B| = \frac{F}{I|L|}\) Substituting the values into the equation: \(|B| = \frac{2.80 \times 10^{-3}}{8.00} = 3.50 \times 10^{-4}\, \text{T}\) Now, we will find the angle between \(\overrightarrow{\mathbf{B}}\) and the Earth's surface in a horizontal plane. Using the given angle of \(58.0^\circ\), we can find the horizontal component of \(\overrightarrow{\mathbf{B}}\) as: \(B_\text{horizontal} = 3.50 \times 10^{-4} \cos{58.0^\circ}\) Now we can find the magnitude of the Earth's magnetic field \(\overrightarrow{\mathbf{B}}\) as: \(|\overrightarrow{\mathbf{B}}| = \frac{B_\text{horizontal}}{\cos{58.0^\circ}} = \frac{3.50 \times 10^{-4}}{\cos{58.0^\circ}} = 6.52 \times 10^{-4}\, \text{T}\)

The direction of the magnetic force on the wire is downward. The magnitude of the Earth's magnetic field \(\overrightarrow{\mathbf{B}}\) is \(6.52 \times 10^{-4}\, \text{T}\).