Question

A 10 cm-long wire carrying a current of 2 A is immersed in a uniform magnetic field of 0.5 T. If the magnetic field is oriented perpendicular to the wire, what is the magnitude of the force that the wire will experience?

Answer 1

The force on a 10 cm-long wire carrying a 2 A current and immersed in a 0.5 T magnetic field oriented perpendicular to it is calculated using the formula F = ILB sin(\(θ\)), resulting in a force of 0.1 N.

Step-by-step explanation:

The student is asking about the force experienced by a current-carrying wire in a magnetic field, according to the formula F = ILB sin(\(θ\)), where F is the force in newtons, I is the current in amperes, L is the length of the wire in meters, B is the magnetic field in teslas, and \(θ\) is the angle between the wire and the magnetic field. In our case, the wire is 10 cm long, which is 0.10 meters, the current is 2 A, the magnetic field is 0.5 T, and the angle is 90 degrees since the field is oriented perpendicular to the wire.

The magnitude of the force that the wire will experience is 0.1 N.

Answer 2

The magnitude of the force experienced by a wire carrying a current in a magnetic field can be determined using the equation F = BIL, where F is the force, B is the magnetic field strength, I is the current, and L is the length of the wire.

Step-by-step explanation:

The magnitude of the force experienced by a wire carrying a current in a magnetic field can be determined using the equation:

F = BIL

where F is the force, B is the magnetic field strength, I is the current, and L is the length of the wire.

In this case, the wire is 10 cm long (0.1 m) and carries a current of 2 A, while the magnetic field is 0.5 T and perpendicular to the wire.

Substituting the given values into the equation, we have:

F = (0.5 T) * (2 A) * (0.1 m)

F = 0.1 N

Therefore, the magnitude of the force that the wire will experience is 0.1 N.