A 0.50m long wire is carrying a current of 8.0A toward the east. The wire passes through a magnetic field of 0.40T toward the north. What is the force which acts on the wire? What is the direction of the force.

To calculate the force acting on the wire, we can use the formula for the magnetic force on a moving charged particle:

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

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

In this case, the current in the wire is 8.0A toward the east, the magnetic field strength is 0.40T toward the north, and the length of the wire is 0.50m. Since the wire passes through the magnetic field, the angle between the current direction and the magnetic field direction is 90 degrees.

Plugging in these values into the formula:

F = 8.0A * 0.40T * 0.50m * sin(90 degrees).

The sine of 90 degrees is equal to 1, so the equation becomes:

F = 8.0A * 0.40T * 0.50m * 1.

Therefore, the force acting on the wire is:

F = 8.0A * 0.40T * 0.50m = 1.60 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), and your fingers in the direction of the magnetic field (north), then the force will be perpendicular to both and will point either upward or downward.

In this case, the force will be directed upward when looking from above the wire.

The force magnitude is B*I*L and the direction is perpendicular to both the I and current directions. In this case, that would be UP. (the direction of the cross product I x B -- use the right hand rule. )