AP Physics 2 Exam Question

A wire carrying a current is placed in a uniform magnetic field. The wire is positioned parallel to the magnetic field lines. The wire has a length of 0.5 meters and carries a current of 2.5 Amperes. The magnetic field strength is 0.15 Tesla.

(a) Calculate the magnetic force acting on the wire.

(b) If the wire is rotated so that it makes an angle of 60 degrees with the magnetic field lines, calculate the magnetic force acting on the wire.

(c) Explain why the magnetic force is zero when the wire is perpendicular to the magnetic field lines.

Answer and Explanation

(a) To calculate the magnetic force acting on the wire, we will use the formula:

where:

• $F$ is the magnetic force,
• $I$ is the current in the wire,
• $L$ is the length of the wire,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the wire and the magnetic field lines.

Given:

• $I = 2.5$ Amperes,
• $L = 0.5$ meters,
• $B = 0.15$ Tesla,
• $\theta = 0^\circ$ (since the wire is parallel to the magnetic field lines).

Solving for $F$:

Since $\sin(0^\circ) = 0$, the magnetic force acting on the wire is $\boxed{0 \, \text{N}}$.

(b) Now, let's calculate the magnetic force when the wire is rotated to an angle of $60^\circ$.

Given:

• $I = 2.5$ Amperes,
• $L = 0.5$ meters,
• $B = 0.15$ Tesla,
• $\theta = 60^\circ$.

Solving for $F$:

Using the trigonometric identity $\sin(60^\circ) = \frac{\sqrt{3}}{2}$:

Simplifying: