Two parallel wires are carrying 5A and are separated by 2.0 m. if the currents are in opposite directions, what is the magnitude of the magnetic field halfway between them?
Someone please help as I can't figure out the question.
Thanks,
They are asking for the field between the wires, not the force.
Because of symmetry, the magnetic field due to each wire will be equal and in the same direction at any point midway between them. (If the currents were parallel, the fields would cancel there and the net would be zero). Just double the field due to one wire. That gives you
B = 2 * muo * I/(2 pi R)
where R = 1.0 m, a muo is the peremeability of free space.
To find the magnitude of the magnetic field halfway between the two wires, you can use the formula:
B = 2 * μ₀ * I / (2 * π * R)
where B is the magnetic field, μ₀ is the permeability of free space (μ₀ = 4π x 10^-7 T*m/A), I is the current in the wire, and R is the distance between the wires.
In this case, the current in each wire is 5A and the distance between the wires is 2.0m. Plugging these values into the formula, you get:
B = 2 * (4π x 10^-7 T*m/A) * 5A / (2 * π * 1.0m)
Now, let's simplify the equation:
B = (4π x 10^-7 T*m/A) * 5A / (2π * 1.0m)
The π's cancel out:
B = (4 x 10^-7 T*m/A) * 5A / (2 * 1.0m)
B = 2 x 10^-7 T
So, the magnitude of the magnetic field halfway between the two wires is 2 x 10^-7 Tesla (T).