A 60 cm long wire is oriented from east to west directly above the equator and carries a current of 5 A (toward the east). What is the magnitude and direction of the magnetic force on this wire due to the earth’s magnetic field (30 µT oriented south and north)?

To calculate the magnitude and direction of the magnetic force on the wire due to the Earth's magnetic field, you can use the formula:

F = I * L * B * sin(theta)

where:
F is the magnetic force,
I is the current flowing in the wire,
L is the length of the wire,
B is the magnetic field strength,
and theta is the angle between the direction of the current and the magnetic field.

In this case, the current is 5 A, the length of the wire is 60 cm (0.6 m), and the magnetic field strength is 30 µT (30 * 10^-6 T).

The wire is oriented from east to west, which means the current is flowing in the east direction. The magnetic field is oriented south and north, which is perpendicular to the flow of current, resulting in a 90-degree angle (theta) between the current and the magnetic field.

Now let's calculate the force:

F = (5 A) * (0.6 m) * (30 * 10^-6 T) * sin(90°)

sin(90°) is equal to 1, so the equation becomes:

F = (5 A) * (0.6 m) * (30 * 10^-6 T) * 1

F = 0.009 N

The magnitude of the magnetic force on the wire is approximately 0.009 N.

The direction of the force can be determined using the right-hand rule: If you point your right thumb in the direction of the current (east to west) and your fingers in the direction of the magnetic field (south to north), your palm will face downward. This means the force on the wire is directed downward, or southward in this case.

Therefore, the magnitude of the magnetic force on the wire is 0.009 N and it is directed southward.