Question

A long straight horizontal wire carries a current I = 2.60 A to the left. A positive 1.00 C charge moves to the right at a distance 1.50 m above the wire at constant speed v = 4500 m/s . (Figure 1) What are the magnitude and the direction of the magnetic force on the charge?

PART A: What is the magnitude B of the magnetic field at the location of the charge due to the current-carrying wire

Answer 1

The magnitude B of the magnetic field at the location of the charge due to the current-carrying wire is 2.18 * 10^-7 T

To find the magnitude B of the magnetic field at the location of the charge due to the current-carrying wire, we can use the formula for the magnetic field created by a long straight wire, which is given by the equation B = (μ0 * I) / (2π * r), where μ0 is the permeability of free space, I is the current, and r is the distance from the wire.

In this case, the current I = 2.60 A and the distance from the wire r = 1.50 m. We can use the given values to calculate the magnitude of the magnetic field.

Substituting the values into the formula, we have B = (4π * 10^-7 T·m/A * 2.60 A) / (2π * 1.50 m). Simplifying this equation gives us B = (4π * 10^-7 T·m/A * 2.60 A) / (3.00 m).

Using the given current of 2.60 A and the distance of 1.50 m, we can calculate the magnitude of the magnetic field. Plugging these values into the formula B = (μ0 * I) / (2π * r), we can simplify the equation to B = (4π * 10^-7 T·m/A * 2.60 A) / (3.00 m). Evaluating this expression gives us B = 2.18 * 10^-7 T. Therefore, the magnitude of the magnetic field at the location of the charge is 2.18 * 10^-7 T.

The magnitude B of the magnetic field at the location of the charge due to the current-carrying wire is 2.18 * 10^-7 T.