Two long electrical distribution lines run parallel to the ground and are separated horizontally by a distance d= 4.5 m. They each carry a current of i= 550 A, with the easternmost line running due north and the westernmost line running due south. Earth’s magnetic field in this area has a magnitude of B= 40μT, and it points due north and at an angle of θ= 7.5°below the horizontal (i.e., towards the ground). What is the magnitude of the net magnetic force on each meter-long section of the eastern line? Hint: there are two magnetic forces acting on each line.
To find the magnitude of the net magnetic force on each meter-long section of the eastern line, we need to calculate the magnetic force for each section and then add them up.
The magnetic force on a current-carrying wire can be calculated using the formula:
F = i * L * B * sin(θ)
Where:
F is the magnetic force,
i is the current,
L is the length of the wire,
B is the magnetic field, and
θ is the angle between the wire and the magnetic field.
In this case, the length of each section of the eastern line is 1 meter, the current is 550 A, the magnetic field is 40μT, and the angle θ is 7.5°.
Using the given information, we can substitute these values into the formula and calculate the magnetic force for each section. Then, we can add up the forces for all the sections of the eastern line to find the net magnetic force.
Let's calculate the magnetic force for each section:
F1 = i * L * B * sin(θ)
= 550 A * 1 m * 40μT * sin(7.5°)
= 550 * 1 * 40 * 10^(-6) * sin(7.5°)
≈ 1.24 * 10^(-3) N
Since the length of each section is 1 meter, the magnetic force on each section of the eastern line is approximately 1.24 * 10^(-3) N.
Therefore, the magnitude of the net magnetic force on each meter-long section of the eastern line is approximately 1.24 * 10^(-3) N.