AP Physics 2 Exam Question:

A particle with a charge of +2e is moving in a uniform magnetic field $\vec{B}$ with a velocity of 10 m/s. If the magnetic field has a magnitude of 0.5 T and is directed perpendicular to the velocity of the particle, what is the force experienced by the particle?

A) 1 N B) 2 N C) 3 N D) 4 N

Answer:

The force experienced by a charged particle moving through a magnetic field can be determined using the formula:

Where:

• $\vec{F}$ is the force experienced by the particle (in Newtons)
• $q$ is the charge of the particle (in Coulombs)
• $\vec{v}$ is the velocity vector of the particle (in m/s)
• $\vec{B}$ is the magnetic field vector (in Tesla)

In this case, the charge of the particle ($q$) is +2e (where $e$ is the elementary charge), the velocity of the particle ($\vec{v}$) is 10 m/s, and the magnetic field ($\vec{B}$) has a magnitude of 0.5 T. Since the magnetic field is directed perpendicular to the velocity of the particle, the angle between $\vec{v}$ and $\vec{B}$ is 90 degrees.

Substituting the given values into the formula, we have:

Since $\sin 90° = 1$, the equation simplifies to:

The elementary charge ($e$) is equal to $1.6 \times 10^{-19} \, \text{C}$.

Evaluating the expression, we find:

Hence, the force experienced by the particle is $\boxed{\text{8 N}}$.