A vertical straight wire carrying an upward 35 current exerts an attractive force per unit length of 8.0×10−4 on a second parallel wire 7.5 away.
What current (magnitude and direction) flows in the second wire?
To determine the current in the second wire, we can use the formula for the Magnetic Force between two parallel wires:
F = (μ0 * I1 * I2 * l) / (2π * d)
Where:
F is the force per unit length between the wires.
μ0 is the permeability of free space (4π × 10^-7 T m/A).
I1 and I2 are the currents in the first and second wires, respectively.
l is the length of the wires in meters.
d is the distance between the wires in meters.
From the given information, we know that:
F = 8.0 × 10^-4 N/m (attractive force per unit length),
I1 = 35 A (current in the first wire),
l = 1 m (since we are considering the force per unit length).
d = 7.5 m (distance between the wires).
Now, let's rearrange the formula to solve for I2:
I2 = (2π * d * F) / (μ0 * I1 * l)
Substituting the given values into the equation:
I2 = (2π * 7.5 * 8.0 × 10^-4) / (4π × 10^-7 * 35 * 1)
Simplifying the expression:
I2 = (15π * 8.0 × 10^-4) / (4π × 10^-7 * 35)
Canceling out the common factors:
I2 = (15 * 8.0 × 10^-4) / (35)
Calculating the result:
I2 ≈ 0.00343 A
Therefore, the magnitude of the current flowing in the second wire is approximately 0.00343 amperes. Since the first wire is carrying an upward current, the second wire will also carry a current in the same direction (upward).