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 find the magnitude and direction of the magnetic force on the wire, you can use the formula:
F = I * L * B * sin(θ)
Where:
F is the magnetic force on the wire,
I is the current in the wire,
L is the length of the wire,
B is the magnetic field strength, and
θ is the angle between the direction of the current and the magnetic field.
In this case, the current is 5 A, and the length of the wire is 60 cm (or 0.6 meters). The magnetic field strength is given as 30 µT (or 30 * 10^(-6) Tesla), which is oriented in the south-north direction.
Since the wire is oriented from east to west, and the magnetic field is oriented south-north, the angle between the current and the magnetic field is 90 degrees.
Now, let's plug in the values into the formula:
F = (5 A) * (0.6 m) * (30 * 10^(-6) T) * sin(90°)
The sine of 90 degrees is 1, so the formula becomes:
F = (5 A) * (0.6 m) * (30 * 10^(-6) T) * 1
Simplifying the expression:
F = 0.009 N
So, the magnitude of the magnetic force on the wire is 0.009 N.
Now, let's determine the direction. According to the right-hand rule, if the current is traveling from east to west and the magnetic field is oriented south to north, the magnetic force will act downward.
Therefore, the direction of the magnetic force on the wire is downward.