Question

A small aluminum ball with a mass of \(5.00 \mathrm{~g}\) and a charge of \(15.0 \mathrm{C}\) is moving northward at \(3000 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.00 \mathrm{~m}\), in a clockwise sense when viewed from above. Ignoring gravity, what is the magnitude and the direction of the magnetic field that must be applied to the aluminum ball to cause it to have this motion?

Short answer

Answer: The magnitude of the applied magnetic field is 0.50 T, and the direction is downward (southward).

Step-by-step solution

Analyze the conditions and forces involved

The aluminum ball needs to travel in a horizontal circular path. The only force acting on it is the magnetic force, which also provides the centripetal force for circular motion.

Write down the magnetic force equation and centripetal force equation

The magnetic force acting on a charged particle can be described by the Lorentz force equation: $$F_B = qvB\sin(\theta)$$ where \(F_B\) is the magnetic force, \(q\) is the charge of the particle, \(v\) is its velocity, \(B\) is the magnetic field, and \(\theta\) is the angle between the velocity and magnetic field vectors. The centripetal force needed for circular motion is given by: $$F_c = \frac{mv^2}{r}$$ 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.

Equate magnetic and centripetal forces

Since the magnetic force provides the centripetal force for circular motion, we can set the two forces equal to each other: $$qvB\sin(\theta) = \frac{mv^2}{r}$$

Solve for the magnitude of the magnetic field

We are asked to find the magnitude \(B\) of the applied magnetic field. Rearrange the equation from the previous step to solve for \(B\): $$B = \frac{mv^2}{rqv\sin(\theta)}$$ Plug in the given values for mass (\(m = 5.00 \mathrm{~g} = 0.005 \mathrm{~kg}\)), charge (\(q = 15.0 \mathrm{C}\)), velocity (\(v = 3000 \mathrm{~m} / \mathrm{s}\)), and radius (\(r = 2.00 \mathrm{~m}\)). Since the ball is supposed to travel clockwise when viewed from above, the magnetic field should be pointing downward (southward), and the angle between the magnetic field and the velocity vector is \(\theta = 90^\circ\). Therefore, \(\sin(\theta) = 1\). $$B = \frac{(0.005 \mathrm{~kg})(3000 \mathrm{~m} / \mathrm{s})^2}{(2.00 \mathrm{~m})(15.0 \mathrm{C})(3000 \mathrm{~m} / \mathrm{s})(1)}$$ $$B = \frac{(0.005)(9\cdot 10^6)}{(2)(15)(3\cdot 10^3)}$$ $$B = \frac{45\cdot 10^3}{90\cdot 10^3}$$ $$B = 0.50 \mathrm{T}$$

Determine the direction of the magnetic field

As mentioned earlier, for the aluminum ball to travel clockwise when viewed from above, the magnetic field should be pointing downward (southward).

Write the final answer

Therefore, the magnitude and direction of the magnetic field that must be applied to the aluminum ball for the desired motion are \(0.50 \mathrm{T}\) and downward (southward), respectively.

Key concepts

Centripetal Force

Centripetal force is a fundamental concept when it comes to understanding circular motion. It is the force required to make an object follow a curved path. Imagine swinging a ball on a string in a circular path; the tension in the string provides the centripetal force pulling the ball towards the center of the circle.
For any object undergoing uniform circular motion, the centripetal force is given by the formula:

• \( F_c = \frac{mv^2}{r} \)

Here, \(m\) is the mass, \(v\) is the velocity, and \(r\) is the radius of the circle. The centripetal force always points towards the center, keeping the object moving in a circle. Without this force, the object would continue in a straight line due to inertia.

• If the speed, mass, or radius changes, the required centripetal force changes as well.
• This concept applies to many real-world situations like cars taking a turn on a highway or planets orbiting a star.

Lorentz Force

The Lorentz force is a key idea when examining how charged particles interact with magnetic fields. It is the force experienced by a charged particle, like our aluminum ball, moving through a magnetic field.
The Lorentz force is described by the equation:

• \( F_B = qvB\sin(\theta) \)

Where:

• \(F_B\) is the magnetic force.
• \(q\) is the charge of the particle.
• \(v\) is the velocity.
• \(B\) is the magnetic field strength.
• \(\theta\) is the angle between the velocity and the magnetic field.

The sine term indicates that the force is strongest when the motion is perpendicular to the field (i.e., \(\theta = 90^\circ\)).
The Lorentz force acts perpendicular to both the velocity and the magnetic field, often resulting in circular or helical paths for charged particles. This is how the charged ball moves in a circle as seen in the problem example.

Circular Motion

Circular motion is an important concept that covers any motion where an object moves along a circular path. This happens frequently in physics, from atoms in a particle accelerator to celestial bodies in space. Circular motion requires a force acting towards the center of the circle, which is known as the centripetal force that we discussed earlier.
There are two types of circular motion:

• Uniform Circular Motion: The object moves at a constant speed along the circular path. The velocity direction changes but magnitude remains constant.
• Non-Uniform Circular Motion: The speed varies at different points in the path.

In uniform circular motion:

• The speed is constant, but velocity (a vector) is constantly changing direction.
• This change in velocity direction means there is acceleration, even if speed is unchanged.
• The centripetal acceleration \(a_c\) is calculated by \(a_c = \frac{v^2}{r}\).

Understanding circular motion is essential for analyzing systems ranging from simple pendulums to complex astronomical systems.