Problem 63 - University Physics with Modern Physics

Question

A slender rod, 0.240 \(\mathrm{m}\) long, reates with an angular speed of 8.80 \(\mathrm{rad} / \mathrm{s}\) about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of 0.650 \(\mathrm{T}\) (a) What is the induced emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at 8.80 \(\mathrm{rad} / \mathrm{s}\) about an axis through its center and perpendicular to the rod, In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?

Step-by-Step Solution

Answer

(a) 0.164 V; (b) 0.164 V; (c) 0.082 V; 0.041 V for center to one end.

Step 1: Understanding the problem

We need to calculate the induced emf in the rotating rod and the potential difference between its ends for two different cases. In the first scenario, the rod rotates about an axis through one end. In the second, it rotates about its center.

Step 2: Calculating emf for the rod rotating about one end

The formula for the induced emf () in a rod rotating about one end is given by = \(\frac{1}{2} B \omega L^2\) where \( B \) is the magnetic field, \( \omega \) is the angular speed, and \( L \) is the length of the rod.Substitute the values:\[ \begin{align*} \varepsilon & = \frac{1}{2} \times 0.650 \times 8.80 \times (0.240)^2\ & = \frac{1}{2} \times 0.650 \times 8.80 \times 0.0576\ & \approx 0.164 V. \end{align*} \]

Step 3: Potential difference for the rod rotating about one end

The potential difference between the ends of the rod when it rotates about one end equals the induced emf. Therefore, \( V = 0.164 \, V \).

Step 4: Calculating potential difference for the rod rotating about its center

When the rod rotates about its center, the induced emf is calculated slightly differently; it creates two emf values on each half of the rod. The potential difference for one half of the rod is given by:\[ \varepsilon_{\text{half}} = \frac{1}{2} B \omega (\frac{L}{2})^2 \approx \frac{1}{2} \times 0.650 \times 8.80 \times \left(\frac{0.240}{2}\right)^2 \]Calculate:\[ \varepsilon_{\text{half}} = \frac{1}{2} \times 0.650 \times 8.80 \times 0.0144 = 0.041 V \]The potential difference between the ends of the rod is twice this value: \[ V = 2 \times 0.041 = 0.082 \, V \].

Step 5: Potential difference between the center and one end

Since the entire difference is partitioned equally across both halves of the rod when rotating about the center, the potential difference between the center and one end is also given by one half of the full potential difference:\( V = 0.041 \, V \).

Key Concepts

Magnetic Field Interaction

Whenever a magnetic field interacts with a moving conductor, such as a rotating rod, an electric potential is induced. This is a direct consequence of Faraday's Law of Induction, which states that a change in magnetic flux through a circuit induces an electromotive force (EMF). In this case, the rotating rod is the moving part in the magnetic field, causing this change in flux.

The magnetic field, denoted by \(B\), is uniform and acts perpendicular to the plane of rotation, enhancing the induction effect. The strength of the induced EMF can be quantified using the formula \(\varepsilon = \frac{1}{2}B\omega L^2\). Here, \(\omega\) represents angular velocity and \(L\) is the length of the rod.

This setup exemplifies how moving conductors in magnetic fields can be utilized to create electric currents, which is the principle behind many power generation technologies.

Rotational Motion

Rotational motion plays a crucial role in this problem. The rod's angular velocity, \(\omega\), at 8.80 \(\text{rad/s}\), signifies how fast the rod spins around a fixed axis. When discussing rotation about one end of the rod, we consider how the entire length of the rod sweeps through space.

The rotational motion about one end results in one side of the rod moving faster than the center, causing a large change in speed across its length. This results in a higher induced EMF compared to rotating about the center. When the rod rotates about its center, both halves move at identical speeds.

Understanding how EMF is induced in rotating systems provides an insight into mechanical systems that transform rotational movement into electricity. The position of the axis in relation to the center of mass can significantly affect the potential difference created.

Physics Problem Solving

To solve physics problems such as this one, it's important to systematically break down the tasks using known formulas and careful observations.

Start by identifying key variables and the scenario details: the magnetic field (\(B\), given as \(0.650\ \text{T}\)), the rod's length (\(L\)), and its angular speed (\(\omega\)). Analyzing these elements helps in applying appropriate physics laws, such as Faraday’s Law, to derive solutions.

The problem requires considering two rotational scenarios—about one end and about the center. Each case uses specific formulas for EMF and potential difference, reflecting the physical arrangement's effect on the outcome.

• For rotation about one end, use \(\varepsilon = \frac{1}{2} B \omega L^2 \).
• For the center, calculate EMF for half and double to get total \(V\).

Practicing such problems enhances understanding of electromagnetic induction, providing tools for tackling more complex systems.