Question

A 5.0 m portion of wire carries a current of 4.0 A from east to west. It experiences a magnetic field of 6.0×10−4 T running from south to north. What is the magnitude and direction of the magnetic force on the wire? Responses 2.4×10−2N upward 2.4 times 10 to the power of negative 2 end exponent N upward, 2.4×10−2N downward 2.4 times 10 to the power of negative 2 end exponent N downward, 1.2×10−2N downward 1.2 times 10 to the power of negative 2 end exponent N downward, 1.2×10−2N upward

Answer 1

The magnitude of the magnetic force on the wire is 2.4×10−2 N. The direction of the force is upward, which is determined using the right-hand rule as the current direction is east to west and the magnetic field direction is south to north.

Step-by-step explanation:

To find the magnitude and direction of the magnetic force on the wire, we can use the right-hand rule and the formula F = I * L * B * sin(θ), where F is the force on the conductor, I is the current, L is the length of the conductor within the magnetic field, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.

In this scenario, the wire carries a current of 4.0 A from east to west, and it experiences a magnetic field of 6.0×10−4 T running from south to north. According to the right-hand rule, if you point your thumb in the direction of the current and your fingers in the direction of the magnetic field, your palm will face the direction of the force. Therefore, since the current goes from east to west and the field from south to north, the force will point upwards.

Now, we apply the formula with the values given: F = 4.0 A * 5.0 m * 6.0×10−4 T * sin(90°). This simplifies to F = 4.0 A * 5.0 m * 6.0×10−4 T * 1 (since sin(90°) = 1), and thus F = 2.4×10−2 N. The direction, as we found from the right-hand rule, is upward.