Question

A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?

Short answer

Answer: The magnitude of the magnetic force on the dust particle is 7.2 x 10^-21 N, and the direction is north.

Step-by-step solution

Identify the given variables

We are given: - \(B = 0.30\, \mathrm{T}\) (magnetic field strength directed to the east) - \(q = -8.0 \times 10^{-18}\, \mathrm{C}\) (charge of the particle) - \(v = 0.30 \, \mathrm{cm/s}\) (velocity of the particle moving straight down)

Convert the velocity to SI units

We need to convert the velocity, \(v\), from centimeters per second to meters per second: \(0.30\, \mathrm{cm/s} = 0.0030 \, \mathrm{m/s}\).

Calculate the angle between the velocity vector and the magnetic field vector

The velocity vector is directed straight down (south), and the magnetic field vector is directed to the east. Thus, the angle between these two vectors, denoted as \(\theta\), is \(90^\circ\).

Calculate the magnitude of the magnetic force

Using the formula \(F_B = qvB \cdot sin\theta\), we can plug in the values: \(F_B = (-8.0 \times 10^{-18}\, \mathrm{C}) (0.0030\, \mathrm{m/s}) (0.30\, \mathrm{T}) \cdot sin(90^\circ)\) \(F_B = 7.2 \times 10^{-21}\, \mathrm{N}\).

Determine the direction of the magnetic force

According to the right-hand rule, if we align our right hand such that the fingers point in the direction of the magnetic field vector (to the east) and then curl our fingers in the direction of the velocity vector (downward), our thumb will point in the direction of the magnetic force acting on the negatively charged particle. Thus, the magnetic force is acting in the northerly direction. In conclusion, the magnitude of the magnetic force on the dust particle is \(7.2 \times 10^{-21}\, \mathrm{N}\), and the direction is north.