Question:

A long straight wire carrying a current of 5 A is placed in a uniform magnetic field of magnitude 0.3 T. The wire is oriented perpendicular to the magnetic field. Determine the magnitude and direction of the magnetic force acting on a 2 meter long segment of the wire.

Answer:

The magnetic force experienced by a current-carrying wire in a magnetic field can be calculated using the formula:

F = I * L * B * sinθ

where:

• F is the magnetic force (in Newtons)
• I is the current (in Amperes)
• L is the length of the wire segment (in meters)
• B is the magnetic field strength (in Tesla)
• θ is the angle between the current direction and the magnetic field

Given:

• I = 5 A (current)
• L = 2 m (length of the wire segment)
• B = 0.3 T (magnetic field)
• The wire is oriented perpendicular to the magnetic field, which means θ = 90°.

Now, substituting the given values into the formula:

F = (5 A) * (2 m) * (0.3 T) * sin(90°)

Since sin(90°) = 1, the equation simplifies to:

F = (5 A) * (2 m) * (0.3 T) * 1

Simplifying further:

F = 3 N

Therefore, the magnitude of the magnetic force acting on the 2-meter long wire segment is 3 Newtons. Since the wire carries a current in the same direction as the magnetic field is pointing, the force is perpendicular to both the current and the magnetic field.

The direction of the magnetic force can be determined using the right-hand rule. By pointing the thumb of your right hand in the direction of the current (from the positive end to the negative end), and curling your fingers towards the direction of the magnetic field (from north to south poles), your palm will face the direction of the force acting on the wire, i.e., into the screen.

Thus, the direction of the magnetic force on the wire segment is into the screen.