Question

A particle with charge \(q\) is at rest when a magnetic field is suddenly turned on. The field points in the \(z\) -direction. What is the direction of the net force acting on the charged particle? a) in the \(x\) -direction c) The net force is zero. b) in the \(y\) -direction d) in the \(z\) -direction

Short answer

a) A perpendicular force b) A force in the direction of the magnetic field c) The net force is zero d) Cannot be determined Answer: c) The net force is zero.

Step-by-step solution

Recall the Lorentz force equation

The Lorentz force equation is given by $$\vec{F} = q(\vec{v}\times\vec{B}),$$ where \(\vec{F}\) is the net force on the charged particle, q is the charge of the particle, \(\vec{v}\) is the velocity of the particle, and \(\vec{B}\) is the magnetic field.

Determine the particle's velocity

The exercise tells us that the particle is initially at rest. Therefore, the velocity of the particle is \(\vec{v} = 0\).

Calculate the net force on the particle

Now that we know the particle's velocity and the magnetic field direction (in the z-direction), we can plug this information into the Lorentz force equation: $$\vec{F} = q(\vec{v}\times\vec{B}) = q(0\times\vec{B}) = 0.$$ The cross product of the zero vector and any other vector is always the zero vector, so the net force acting on the charged particle is zero. Therefore, the correct answer is: c) The net force is zero.