Question:

A wire carrying a current, I, is placed in a uniform magnetic field, B. The wire is oriented perpendicular to the magnetic field lines. The length of the wire is L and its resistance is R. Determine the magnitude of the force acting on the wire in terms of I, B, L, and R.

Answer:

To determine the magnitude of the force acting on the wire, we will use the formula for the magnetic force on a current-carrying wire in a uniform magnetic field:

F = I L B sin(theta),

where F is the force acting on the wire, I is the current flowing through the wire, L is the length of the wire, B is the magnetic field strength, and theta is the angle between the direction of the current and the magnetic field lines.

In this case, since the wire is oriented perpendicular to the magnetic field lines, the angle theta is 90 degrees, which means sin(theta) equals 1.

Substituting the given values into the formula, we get:

F = I L B sin(90°)

F = I L B

The magnitude of the force acting on the wire in terms of I, B, and L is I L B.

However, we are also given the resistance of the wire, R. We can use Ohm's Law, V = I R, to determine the voltage across the wire.

Since Power = Voltage * Current, we have:

Power = (I R) I = I^(2) R.

The power dissipated by the wire is the product of the current squared and the resistance. This power is equal to the force (F) times the velocity (v), since Work = Force * Distance = Power * Time.

Therefore, we can write:

F L = I^(2) R.

Simplifying this expression, we get:

F = (I^(2) R) / L.

Hence, the magnitude of the force acting on the wire in terms of I, B, L, and R is (I^(2) R) / L.