Two long wires, each carrying a current of 9.6A, lie a distance of 49cm from each other. What is the magnetic force per unit length exerted by one on the other.
Answer is in N
You do not specify the directions so we can only say the magnitude
49 cm = 0.49 meters
F = uo i*i / 2 pi d) = uo i^2/(2 pi d) = (4 pi*10^-7) 9.6^2 / (2 pi * .49)
from the Biot-Savart law, the magnetic field from one wire is
B=mu*I/(2PI*d) where I is 9.6, d = .49m, you know PI and mu
now the force jper unit length on the wire which cuts that magnetic field is
Force/length=I*B=9.6*B, where B is above.
To find the magnetic force per unit length exerted by one wire on the other, we can use the formula for the magnetic force between two parallel wires. The formula is given by:
F = (μ₀ * I₁ * I₂ * L) / (2π * d),
where:
F is the magnetic force,
μ₀ is the permeability of free space (μ₀ = 4π * 10^-7 T·m/A),
I₁ and I₂ are the currents in the two wires (in this case, both currents are 9.6A),
L is the length of the wires, and
d is the distance between the wires (in this case, 49cm).
Let's plug in the given values into the formula:
F = (4π * 10^-7 T·m/A) * (9.6A) * (9.6A) * L / (2π * 49cm).
Now, we need to convert centimeters to meters, as the SI units are used in the formula:
F = (4π * 10^-7 T·m/A) * (9.6A) * (9.6A) * L / (2π * 0.49m)
F = (4π * 10^-7 T·m/A) * (9.6A) * (9.6A) * L / (0.98m).
Next, we can simplify the equation further:
F = (4π * 10^-7 T·m/A) * (92.16A²) * L / (0.98m)
F = (3.7699 * 10^-5 T·m) * (92.16A²) * L.
Therefore, the magnetic force per unit length (F) exerted by one wire on the other can be calculated as:
F = (3.7699 * 10^-5 T·m) * (92.16A²) * L.
Note that the answer will depend on the values of L and the units used for L. If you have a specific value for L, you can substitute it into the equation to find the force.