Circular Motion of Charges in Magnetic Fields: Videos & Practice Problems

When a charged particle moves through a magnetic field, it experiences a magnetic force that is always perpendicular to its velocity, resulting in circular motion. The centripetal force required for this motion can be expressed as Fb = mv^2r. This leads to the important equation for radius: r = mvqB. If the charge moves parallel to the field, it continues in a straight line, while at an angle, it exhibits helical motion. The work done by the magnetic force is zero, as it is always perpendicular to the displacement.

Circular Motion of Charges in Magnetic Fields

When a charged particle moves through a magnetic field, it experiences a magnetic force that is always perpendicular to its velocity. This perpendicular relationship results in circular motion, as the force continuously changes the direction of the particle's velocity without altering its speed. To visualize this, one can use the right-hand rule: point your thumb in the direction of the charge's velocity and curl your fingers in the direction of the magnetic field. The direction your palm faces indicates the direction of the magnetic force acting on the charge.

In the context of circular motion, the magnetic force acts as the centripetal force required to keep the charge moving in a circular path. The fundamental equation governing circular motion is given by Newton's second law, expressed as:

$$F = ma$$

For circular motion, this can be rewritten as:

$$F_{c} = m \frac{v^2}{r}$$

where \(F_{c}\) is the centripetal force, \(m\) is the mass of the particle, \(v\) is its velocity, and \(r\) is the radius of the circular path. In the case of a charged particle in a magnetic field, the magnetic force \(F_{B}\) can be expressed as:

$$F_{B} = qvB \sin(\theta)$$

where \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field. For a charge moving perpendicular to the magnetic field (\(\theta = 90^\circ\)), the sine function equals 1, simplifying the equation to:

$$F_{B} = qvB$$

Setting the magnetic force equal to the centripetal force gives:

$$qvB = m \frac{v^2}{r}$$

By rearranging this equation, one can derive the radius of the circular path:

$$r = \frac{mv}{qB}$$

This equation is crucial for understanding the motion of charged particles in magnetic fields and is often remembered using the mnemonic "momentum quarterback," where momentum \(p = mv\) and \(QB\) represents the charge and magnetic field strength.

For example, if an electron (with charge \(q = -1.6 \times 10^{-19}\) coulombs) enters a magnetic field of strength \(B = 0.2\) Tesla, moving perpendicular to the field, and is observed to follow a circular arc with a radius of \(0.3\) centimeters (or \(0.003\) meters), one can calculate its velocity using the derived formula. The mass of the electron is \(9.1 \times 10^{-31}\) kg. Plugging in the values:

$$v = \frac{r \cdot q \cdot B}{m}$$

Substituting the known values yields a velocity of approximately \(1.1 \times 10^{8}\) m/s, which is less than the speed of light, confirming the calculation's validity.

It is also important to note the behavior of charges in different orientations to the magnetic field. If a charge moves parallel to the magnetic field (\(\theta = 0^\circ\)), the magnetic force is zero, and the charge continues in a straight line. Conversely, if the charge moves at an angle to the magnetic field, it exhibits helical motion, spiraling while moving forward.

Lastly, the work done by the magnetic force is always zero when a charge moves in a circular path. This is because the angle between the force (centripetal) and the displacement (tangential) is \(90^\circ\), leading to:

$$W = F \Delta x \cos(90^\circ) = 0$$

Thus, the work done by any centripetal force is consistently zero, reinforcing the concept that magnetic forces do not perform work on charged particles in circular motion.

Here’s what students ask on this topic:

How does a charged particle move in a magnetic field?

When a charged particle moves through a magnetic field, it experiences a magnetic force that is always perpendicular to its velocity. This perpendicular force causes the particle to undergo circular motion. The centripetal force required for this motion is provided by the magnetic force, which can be expressed as $F_b=\frac{{\mathrm{mv}}^2}{r}$. This leads to the important equation for the radius of the circular path: $r=\frac{\mathrm{mv}}{\mathrm{qB}}$. If the charge moves parallel to the magnetic field, it continues in a straight line, while at an angle, it exhibits helical motion.

What is the equation for the radius of a charged particle's circular motion in a magnetic field?

The radius of a charged particle's circular motion in a magnetic field is given by the equation $r=\frac{\mathrm{mv}}{\mathrm{qB}}$. Here, $m$ is the mass of the particle, $v$ is its velocity, $q$ is the charge, and $B$ is the magnetic field strength. This equation is derived from the balance between the centripetal force and the magnetic force acting on the particle.

What happens to a charged particle moving parallel to a magnetic field?

When a charged particle moves parallel to a magnetic field, it does not experience any magnetic force because the angle between the velocity and the magnetic field is 0 degrees. The magnetic force is given by $F_b=qvB\sin \theta$, and since $\sin 0$ is 0, the force is zero. As a result, the particle continues to move in a straight line without any deflection.

What is helical motion in the context of charged particles in a magnetic field?

Helical motion occurs when a charged particle moves at an angle to a magnetic field. In this case, the particle's velocity has both parallel and perpendicular components relative to the magnetic field. The perpendicular component causes circular motion, while the parallel component causes linear motion along the field lines. The combination of these motions results in a helical or spiral path. The particle moves forward while simultaneously spiraling around the magnetic field lines.

Why is the work done by the magnetic force on a charged particle zero?

The work done by the magnetic force on a charged particle is zero because the magnetic force is always perpendicular to the particle's displacement. Work is given by the equation $W=F\Delta x\cos \theta$, where $\theta$ is the angle between the force and displacement vectors. Since the angle is 90 degrees, $\cos 90$ is 0, making the work done by the magnetic force zero. This is true for any centripetal force acting on a particle in circular motion.