A uniform magnetic field of \(0.50 \mathrm{T}\) is directed to the north. At some instant, a particle with charge \(+0.020 \mu \mathrm{C}\) is moving with velocity \(2.0 \mathrm{m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What is the magnitude of the magnetic force on the charged particle? (b) What is the direction of the magnetic force?

We are given the following information: 1. The magnetic field, \( B = 0.50 \, \text{T}\) 2. The particle charge \( q = +0.020 \times 10^{-6} \, \text{C}\) 3. The velocity of the particle \( v = 2.0 \, \text{m/s}\) 4. The angle between the magnetic field and the velocity vector, \(\theta = 30°\)

The formula for the magnetic force acting on a charged particle is given by: \( F = q v B \sin\theta \) where \(F\) is the magnetic force, \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field, and \(\theta\) is the angle between the velocity vector and the magnetic field. We can plug in the given values to compute the magnetic force: \( F = (0.020 \times 10^{-6} \, \text{C}) \cdot (2.0 \, \text{m/s}) \cdot (0.50 \, \text{T}) \cdot \sin(30°)\) We know that \(\sin(30°) = 0.5\), thus: \( F = (0.020 \times 10^{-6} \, \text{C}) \cdot (2.0 \, \text{m/s}) \cdot (0.50 \, \text{T}) \cdot (0.5)\) \( F = 1.0 \times 10^{-6} \, \text{N} \) The magnitude of the magnetic force acting on the charged particle is \(1.0 \times 10^{-6} \, \text{N}\).

To find the direction of the magnetic force, we can use the right-hand rule. Point your fingers of the right hand in the direction of the particle's velocity (east direction in our case) and curl them to the direction of the magnetic field (north direction). The thumb will point in the direction of the force acting on the particle. The thumb points upward, meaning that the direction of the magnetic force acting on the charged particle is upward.

(a) The magnitude of the magnetic force on the charged particle is \(1.0 \times 10^{-6} \, \text{N}\). (b) The direction of the magnetic force is upward.