Question

A long, straight wire carrying a current of 2.5 A is placed in a uniform magnetic field. The wire is in the shape of a square with sides of length 0.5 m. The magnetic field points into the page and has a magnitude of 0.75 T. Determine the magnetic force acting on one section of the wire.

Answer

The magnetic force exerted on a current-carrying wire in a magnetic field is given by the equation:

F = I * B * L * sin(θ)

Where: F is the magnetic force I is the current in the wire B is the magnetic field L is the length of the wire θ is the angle between the wire and the magnetic field

In this case, we have a square wire with sides of length 0.5 m, carrying a current of 2.5 A. The magnetic field has a magnitude of 0.75 T and points into the page. To find the magnetic force, we first need to determine the length of one section of the wire.

Since the wire is a square, each side (L) of the wire is equal to 0.5 m. Therefore, the length of one section of the wire is also 0.5 m.

The angle between the wire and the magnetic field is 90 degrees since the magnetic field points into the page and the wire lies in the plane of the page. Thus, sin(θ) = sin(90°) = 1.

Now we can substitute the given values into the equation:

F = (2.5 A) * (0.75 T) * (0.5 m) * (1)

F = 0.9375 N

Therefore, the magnetic force acting on one section of the wire is 0.9375 N.