A horizontal power line carries a current of 5000 A from south to north. Earth's magnetic field

A horizontal power line carries a current of 5000 A from...

A horizontal power line carries a current of $5000 \mathrm{~A}$ from south to north. Earth's magnetic field $(60.0 \mu \mathrm{T})$ is directed toward the north and inclined downward at $70.0^{\circ}$ to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on $100 \mathrm{~m}$ of the line due to Earth's field.

Key Concepts

Lorentz Force Law

This principle states that a current-carrying conductor in a magnetic field experiences a force given by F = I(L × B). It quantifies the interaction between the electric current and the magnetic field, linking the magnitude and direction of the current, the length of the conductor, and the applied magnetic flux density.

Magnetic Force on a Current-Carrying Conductor

This concept involves calculating the force exerted on a conductor carrying an electric current when placed in an external magnetic field. The force is determined by the product of the current, the length of the conductor within the field, and the component of the magnetic field perpendicular to the conductor, following the cross product relationship.

Vector Cross Product

The cross product is a mathematical operation used to determine the magnitude and direction of the force in the Lorentz force law. It requires understanding the sine of the angle between the current direction and the magnetic field vector, and its result is a vector perpendicular to both, which is essential for correctly determining the force's orientation.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the vector resulting from the cross product. When dealing with magnetic forces on conductors, it helps to physically determine the orientation of the force relative to the direction of the current and the magnetic field.

Magnetic Field Inclination

Magnetic field inclination refers to the angle at which the magnetic field deviates from the horizontal. When calculating forces, only the component of the magnetic field perpendicular to the conductor contributes to the force, making it important to resolve the magnetic field into its horizontal and vertical components based on the provided inclination angle.

Transcript

00:01 Okay, folks, in this video, we're going to be talking about this problem.

00:04 A horizontal power line carries a current of 5 ,000 emperors from south to north.

00:10 Earth's magnetic field is directed towards the earth and inclined downward to 70 degrees to the horizontal.

00:17 We're going to find the magnitude and direction of the magnetic force on 100 meters of the line due to the earth's field.

00:25 Okay, so let me just draw this problem out for you.

00:30 So if this is, the vertical axis is, you know, this is, this is the north, this is the south, and you have a power line that is carrying current, you know, the current goes this way, and you know that the earth's magnetic field is directed towards the north.

00:54 Let me use a different color for that, towards the north, and inclined downward at, that doesn't look like 70 degrees.

01:03 This looks like 70 degrees.

01:05 Okay, so this is 70 degrees to the horizontal.

01:11 North, south, east, west.

01:15 So this is the west.

01:17 This is the east.

01:18 Okay.

01:20 So the earth's magnetic field points this way, and the current points this way.

01:27 Okay.

01:29 So how do we find the magnitude of the magnetic force? on this current carrying wire, well, if you remember, if you have a charged particle, dq, and it's moving in a magnetic field, then that force, then the force that the particle experiences is going to be given by this formula, df equals dq, b across b.

01:53 Okay, and, but we have, in this problem, we have more than just a particle.

01:57 We have, you know, a lot of particles.

01:59 We have a current carrying wire.

02:01 We have a lot of particles in, this wire, a lot of electrons that are moving.

02:06 So we're going to do an integral.

02:08 We're going to sum over all of the particles that are moving.

02:11 And we can rewrite this velocity as dl, dt, and then this cross b.

02:20 And then we can move this dq and dt together.

02:24 So dq over dt, but that becomes current...