# Ball rolling in a magnetic field

• etotheipi
In summary, the conversation discusses the Lorentz force on a ball with spherical symmetry, and how it is equivalent to the magnetic force on a point charge at the center of mass. The conversation also applies Euler's law of motion and considers torques about the center of mass, leading to a discussion about the direction of the friction force and magnetic force. It is determined that there is an inconsistency in the signs used, and the correct values for the angular velocity and radius in circular motion are obtained.
etotheipi
Homework Statement
A ball of radius R, mass M and uniform charge Q is set rolling with its centre of mass initially moving at a speed ##V##. The magnetic field is perpendicular to the surface, which is flat and has a large enough coefficient of static friction to prevent slippage. Determine the trajectory of the ball.
Relevant Equations
N/A
I first found the Lorentz force on the ball as a whole$$\vec{F}_m = \iiint_V \rho(\omega \times \vec{r} + \vec{V})\times \vec{B} dV = \rho \vec{\omega} \times \left( \iint_V \vec{r} dV \right) \times \vec{B} + \rho \iiint_V \vec{V} \times \vec{B} dV = Q\vec{V} \times \vec{B}$$due to the spherical symmetry of the ball, which demonstrates that the magnetic force on the ball is equivalent to that on a point charge at the centre of mass. For the rest I will take the ##\hat{z}## direction as upwards.

Euler's law of motion applied to the ball, with the force of static friction ##\vec{F}_f## and the magnetic force ##\vec{F}_m##, yields$$Q\vec{V} \times \vec{B} + \vec{F}_f = m\vec{a}_{cm}$$Then I also took torques about the centre of mass, noting that the magnetic force causes a torque (due to the presence of a magnetic dipole caused by induced currents in the ball) which is ##\vec{\tau}_m = \frac{Q}{2m} \vec{L}_{cm} \times \vec{B} = \frac{Q}{5}R^2 \vec{\omega} \times \vec{B}## where ##\vec{L}_{cm} = \frac{2}{5}mR^2 \vec{\omega}## is the angular momentum about the centre of mass:$$\sum \vec{\tau}_{cm} = (-R\hat{z})\times \vec{F}_f + \frac{Q}{5}R^2 \vec{\omega} \times \vec{B} = \frac{2}{5}mR^2 \vec{\alpha}$$Now, I define the unit vector ##\hat{n}## in the direction of the frictional force such that ##\vec{F}_f = \mu m g \hat{n}## and write ##\vec{\alpha}## as $$\vec{\alpha} = -\frac{5}{2mR} \hat{z} \times (\mu mg) \hat{n} + \frac{Q}{2m}\vec{\omega} \times \vec{B}$$I can now use the rolling condition, ##\vec{a}_{cm} = R\vec{\alpha} \times \hat{z}##, to write down$$\vec{a}_{cm} = -\frac{5\mu g}{2}\hat{n} + \frac{QR}{2m} \omega B \hat{n}$$This demonstrates that the friction force is oppositely directed to the magnetic force. Finally I can insert this into the Euler's law equation to obtain$$-QVB\hat{n} + \mu m g \hat{n} = -\frac{5\mu mg}{2} \hat{n} + \frac{QVB}{2}\hat{n}$$This means that we can rewrite ##\vec{a}_{cm} = -\frac{4}{7} \frac{QVB}{m} \hat{n} = -r \Omega^2 \hat{n}##. However, the solution manual notes that ##\Omega = \frac{6QB}{7m}## and ##r = \frac{7mV}{6QB}## for the resulting circular motion, which isn't what I get. I'm out on ##\Omega## by a factor of ##\frac{2}{3}##. So I wondered whether anyone could see where I have gone wrong? Thanks!

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I think your work is basically good. There is no need to write the friction force in terms of the coefficient of friction. In fact, the friction is static friction. So, it is not necessarily true that ##f = \mu mg##. In general, ##f \leq \mu mg##.

etotheipi said:
Now, I define the unit vector ##\hat{n}## in the direction of the frictional force such that ##\vec{F}_f = \mu m g \hat{n}## and write ##\vec{\alpha}## as $$\vec{\alpha} = -\frac{5}{2mR} \hat{z} \times (\mu mg) \hat{n} + \frac{Q}{2m}\vec{\omega} \times \vec{B}$$I can now use the rolling condition, ##\vec{a}_{cm} = R\vec{\alpha} \times \hat{z}##, to write down$$\vec{a}_{cm} = -\frac{5\mu g}{2}\hat{n} + \frac{QR}{2m} \omega B \hat{n}$$ This demonstrates that the friction force is oppositely directed to the magnetic force.
I believe there is an inconsistency in your signs here. Note that ##(\vec \omega \times \vec B) \times \hat z## points in the direction of the magnetic force. But you took ##(\vec \omega \times \vec B) \times \hat z## to be in the direction of ##\hat n##. You defined ##\hat n## as the direction of the friction force and concluded that the direction of the friction force is opposite to the direction of the magnetic force. So, something's wrong.

etotheipi
TSny said:
I think your work is basically good. There is no need to write the friction force in terms of the coefficient of friction. In fact, the friction is static friction. So, it is not necessarily true that ##f = \mu mg##. In general, ##f \leq \mu mg##.

I believe there is an inconsistency in your signs here. Note that ##(\vec \omega \times \vec B) \times \hat z## points in the direction of the magnetic force. But you took ##(\vec \omega \times \vec B) \times \hat z## to be in the direction of ##\hat n##. You defined ##\hat n## as the direction of the friction force and concluded that the direction of the friction force is opposite to the direction of the magnetic force. So, something's wrong.

You're completely right, ##\frac{Q}{2m}(\vec{\omega} \times \vec{B}) \times \hat{z} = -\frac{QVB}{2m} \hat{n}##, and if I substitute that through I get that, along with ##V = r\Omega##,$$-\frac{6}{7}\frac{QVB}{m} = -r\Omega^2$$ $$\Omega = \frac{6}{7}\frac{QB}{m}$$which is what is given in the solutions.

Thank you ☺

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TSny

## 1. What is ball rolling in a magnetic field?

Ball rolling in a magnetic field refers to the phenomenon where a ball or any other object with magnetic properties is placed in a magnetic field and experiences a force that causes it to roll or move in a certain direction.

## 2. How does a magnetic field affect the motion of a rolling ball?

A magnetic field exerts a force on a moving object with magnetic properties, such as a rolling ball. This force, known as the Lorentz force, causes the ball to experience a change in direction, causing it to roll in a curved path.

## 3. What factors affect the movement of a ball rolling in a magnetic field?

The movement of a ball rolling in a magnetic field is affected by several factors, including the strength of the magnetic field, the mass and magnetic properties of the ball, and the angle at which the ball enters the magnetic field.

## 4. Can a ball rolling in a magnetic field be used as a source of energy?

Yes, a ball rolling in a magnetic field can be used as a source of energy through a process called magnetic induction. As the ball moves through the magnetic field, it generates an electric current which can then be harnessed to power devices.

## 5. What are some real-world applications of ball rolling in a magnetic field?

Ball rolling in a magnetic field has several real-world applications, including in generators, motors, and magnetic levitation systems. It is also used in scientific experiments to study the effects of magnetic fields on moving objects.

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