A long, straight horizontal wire carries a current I = 2.10 A to the left. A positive 1.00 C charge moves to the right at a distance of 3.00 m above the wire at a constant speed of v = 4250 m/s. What is the magnitude and the direction of the magnetic force on the charge?What is the magnitude BBB of the magnetic field at the location of the charge due to the current-carrying wire? Express your answer in teslas.

The magnetic force on the charge is 5.95*10^(-4) Newtons and is directed vertically upward. The magnitude of the magnetic field of the current at the location of the charge is 1.4*10^(-7) Tesla.

The magnetic field lines of a long current-carrying wire are concentric circles in the plane perpendicular to the wire. The direction of the field is determined by the right-hand rule.

If the wire is horizontal and the current is to the left, then the magnetic field will be directed into the page at a point above the wire and out of the page at a point under the wire (please see the attached image).

The magnitude of the field at a point distance r away from the wire is

`B = (mu_o*I)/(2*pi*r)`

Here, `mu_0` is the magnetic permeability of vacuum and is `mu_0 = 4*pi*10^(-7) T*m/A`

For the current of I = 2.1 A and the distance r = 3 m, the magnitude of the field is

`B = (4*pi*10^(-7) * 2.1)/(2*pi*3) = 1.4*10^(-7) T`

The magnetic force on the moving charge can be found as

`vec F = q*vec v xx vecB`

Since the charge is positive and is moving to the right, and the magnetic field is pointing into the page, the direction of the force, as determined by the right-hand rule, will be upward. (Again, please see the attached image.) The magnitude of the force will be

`F = qvB = 1*4250*1.4*10^(-7) = 5.95*10^(-4) N`

See the reference link below discussing the magnetic field of currents and illustrating the right-hand rule.