Problem 88 - University Physics with Modern Physics

Question

A circular ring with area \(4.45 \mathrm{~cm}^{2}\) is carrying a current of 12.5 A. The ring is free to rotate about a diameter. The ring, initially at rest, is immersed in a region of uniform magnetic field given by \(\vec{B}=\left(1.15 \times 10^{-2} \mathrm{~T}\right)(12 \hat{\imath}+3 \hat{\jmath}-4 \hat{k}) .\) The ring is positioned initially such that its magnetic moment is given by \(\vec{\mu}_{1}=\mu(-0.800 \hat{\imath}+0.600 \hat{\jmath}),\) where \(\mu\) is the (positive) magnitude of the magnetic moment. The ring is released and turns through an angle of \(90.0^{\circ},\) at which point its magnetic moment is given by \(\vec{\mu}_{f}=-\mu \hat{k} .\) (a) Determine the decrease in potential energy. (b) If the moment of inertia of the ring about a diameter is \(8.50 \times 10^{-7} \mathrm{~kg} \cdot \mathrm{m}^{2},\) determine the angular speed of the ring as it passes through the second position.

Step-by-Step Solution

Answer

(a) Decrease in potential energy: -0.000497 J. (b) Angular speed: 34.25 rad/s.

Step 1: Calculate the Magnitude of the Magnetic Moment

The magnitude of the magnetic moment \( \mu \) is given by the product of the current \( I \) and the area \( A \) of the ring. Thus, \[ \mu = I \times A = 12.5 \times 4.45 = 55.625 \; \text{A} \cdot \text{cm}^2 \]. To convert this to \( \text{A} \cdot \text{m}^2 \), we have \( \mu = 55.625 \times 10^{-4} \; \text{A} \cdot \text{m}^2 = 0.0055625 \; \text{A} \cdot \text{m}^2 \).

Step 2: Express Magnetic Moment Vectors

Utilize the unit vector form to express initial \( \vec{\mu}_1 \) and final \( \vec{\mu}_f \) magnetic moments. \( \vec{\mu}_1 = 0.0055625(-0.800 \hat{\imath} + 0.600 \hat{\jmath}) \) and \( \vec{\mu}_f = -0.0055625 \hat{k} \).

Step 3: Calculate Initial and Final Potential Energy

The potential energy of a magnetic moment in a magnetic field is given by \( U = -\vec{\mu} \cdot \vec{B} \). Calculate \( U_1 \) and \( U_f \) using the dot product:\[ U_1 = -0.0055625(-0.800 \hat{\imath} + 0.600 \hat{\jmath}) \cdot (1.15 \times 10^{-2})(12 \hat{\imath} + 3 \hat{\jmath} - 4 \hat{k}) \]\[ = -0.0055625 \times 1.15 \times 10^{-2} \times ( (-0.800\times12) + (0.600\times3) ) \]\[ = -0.0055625 \times 1.15 \times 10^{-2} \times (-9.6 + 1.8) \]\[ = -0.0055625 \times 1.15 \times 10^{-2} \times (-7.8) = 0.000497 \; \text{J} \]Similarly, \( U_f = 0 \) since \( \vec{\mu}_f \cdot \vec{B} = 0 \) because \( \vec{\mu}_f \) is perpendicular to \( \vec{B} \).

Step 4: Calculate Decrease in Potential Energy

The decrease in potential energy \( \Delta U \) can be calculated as \( \Delta U = U_f - U_1 = 0 - 0.000497 = -0.000497 \; \text{J} \).

Step 5: Use Energy Conservation to Find Angular Speed

The change in potential energy is converted to kinetic rotational energy. Use the formula for rotational kinetic energy which is \( K = \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed. Set \( \Delta U = K \): \[ -0.000497 = \frac{1}{2} \times 8.50 \times 10^{-7} \times \omega^2 \]Solving for \( \omega \) gives:\[ \omega^2 = \frac{-0.000497 \times 2}{8.50 \times 10^{-7}} \]\[ \omega = \sqrt{\frac{0.000994}{8.50 \times 10^{-7}}} \approx 34.25 \; \text{rad/s} \]

Step 6: Verify Conservation of Energy

Confirm that the calculated decrease in potential energy equals the increase in kinetic energy, ensuring energy conservation holds true for the system.

Key Concepts

Magnetic Moment

The magnetic moment is a key concept in electromagnetism. It tells us how strong and in which direction a magnetic object can affect or be affected by a magnetic field. For any loop carrying an electric current, such as a ring, the magnetic moment

• is the product of the current (\( I \)) flowing through the loop and the area (\( A \)) the loop encloses.
• For the example in the exercise, the magnetic moment is \( \mu = I \times A = 12.5 \times 4.45 \; \text{A} \cdot \text{cm}^2 \).

To make units consistent, convert it to \( \text{A} \cdot \text{m}^2 \) by multiplying by \( 10^{-4} \).
In simpler words, bigger the current or larger the loop, the stronger the magnetic moment will be. This moment is a vector quantity, which means it has both magnitude and direction. The orientation of this vector is determined by the right-hand rule and is perpendicular to the plane of the loop.

Potential Energy

In this exercise, potential energy relates to the position of the magnetic moment in a magnetic field. It's defined as the energy due to the orientation of the magnetic moment in an external magnetic field. The formula is:

• \( U = -\vec{\mu} \cdot \vec{B} \)
• The potential energy is at its minimum when the magnetic moment aligns with the magnetic field fully.

The exercise demonstrated how to compute the change in potential energy when the ring rotates under the magnetic field's influence. Initially, the potential energy \( U_1 \) was computed with the magnetic moment aligned at a certain angle with respect to the magnetic field.
Upon rotating by \( 90^{\circ} \), the potential energy drops to zero because the magnetic field and moment are perpendicular and do not interact directly. Such a decrease in potential energy indicates release or gain in energy stored in any magnetic alignment. This problem exemplifies how potential energy conversion can indirectly cause motion.

Angular Speed

Angular speed is the rate at which an object rotates or spins. It describes how quickly an object makes one full rotation. In this exercise, angular speed is significant because the ring rotates due to the transfer of potential energy to kinetic energy. Here's how it's calculated:

• The decrease in potential energy becomes kinetic energy, resulting in rotational motion.
• The kinetic energy \( K \) for rotation is described by: \( K = \frac{1}{2}I\omega^2 \)

To find the angular speed \( \omega \), set the decrease in potential energy equal to kinetic energy. Solving \( \omega^2 = \frac{0.000994}{8.50 \times 10^{-7}} \), where \( I \) is the moment of inertia and \( \omega \) gives the rotational aspect.
This angular speed reflects how quickly the ring rotates after being repositioned in the magnetic field. Fast rotation implies a quickened transformation processes where energy conversion is ideal.