Animated Solution for Physics - Magnetic Effects of Current: A beam of protons with a velocity $4\times 10^5$ m/s enters a uniform magnetic field of $0.3$ T at an angle of $60^{\circ }$ to the magnetic field. Find the radius of the helical path taken by the protons beam. Also find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation).

Imagine you are a proton, fired out of a particle accelerator at a blistering speed of $4\times 10^5\text{ m/s}$. Suddenly, you enter a region permeated by a uniform magnetic field of $0.3\text{ T}$. If you had entered perfectly perpendicular to the field, you would be trapped in an endless, flat circular orbit. If you had entered perfectly parallel, you would feel absolutely no force and continue in a straight line forever.

But in this scenario, you enter at a rebellious angle of $60^{\circ }$. What happens now?

To understand the physics, we must invoke the principle of superposition. We break your velocity vector $\mathbf{v}$ into two independent components: 1. The Perpendicular Component ($v_{\perp }$): This component is perpendicular to the magnetic field. According to the Lorentz force law, $\mathbf{F}=q(\mathbf{v}\times \mathbf{B})$, this component generates a constant force that acts as a centripetal force, whipping you into a circular frenzy. 2. The Parallel Component ($v_{\parallel }$): This component is perfectly aligned with the magnetic field. Because the cross product of parallel vectors is zero, the magnetic field exerts absolutely zero force on this component. It acts like a ghost, carrying you forward at a constant speed.

When you combine a circular motion with a constant forward translation, you trace out a beautiful, three-dimensional helix.

Finding the Radius

Let's calculate the radius of the circular part of your helical journey. The magnetic force provides the centripetal force:

Solving for the radius $r$, we get:

Now, we simply substitute the standard constants for a proton ($m=1.67\times 10^{-27}\text{ kg}$, $q=1.6\times 10^{-19}\text{ C}$) and our given values:

Knowing that $\sin 60^{\circ }=\frac{\sqrt{3}}{2}\approx 0.866$, the arithmetic simplifies beautifully:

Calculating the Pitch

The pitch of a helix is the linear distance you travel forward along the magnetic field during exactly one complete circular rotation. Think of it as the distance between two adjacent threads on a screw.

To find the pitch, we need to know how long one rotation takes. This is the time period $T$. The beauty of cyclotron motion is that the time period is completely independent of your velocity!

During this time $T$, your parallel velocity $v_{\parallel }$ has been carrying you forward. Therefore, the pitch $p$ is:

Let's plug in the numbers, noting that $\cos 60^{\circ }=0.5$:

By simply decomposing a vector, we've completely mapped out a complex three-dimensional trajectory. The proton spirals through the magnetic field with a radius of $1.2\text{ cm}$, advancing forward by $4.37\text{ cm}$ with every single loop. This exact principle is what governs the spectacular auroras at the Earth's poles, where charged particles from the solar wind spiral along the Earth's magnetic field lines!