/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 A thin metal rod of length \(\el... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin metal rod of length \(\ell\) rotates with angular velocity \(\omega\) about an axis through one end (Fig. \(29-51\) ). The rotation axis is perpendicular to the rod and is parallel to uniform magnetic field \(\overrightarrow{\mathbf{B}}\). Determine the emf developed between the ends of the rod.

Short Answer

Expert verified
The emf is \( \frac{1}{2} B \omega \ell^2 \).

Step by step solution

01

Understand the Problem

We need to find the electromotive force (emf) generated in a rotating rod. The rod rotates about one end, and the magnetic field is uniform and parallel to the axis of rotation. As the rod moves through the magnetic field, a voltage (emf) is generated between its two ends.
02

Use Faraday's Law of Induction

According to Faraday's law, an emf is induced in a conductor moving through a magnetic field, given by the formula: \[ \text{emf} = -\frac{d\Phi_B}{dt} \] where \(\Phi_B\) is the magnetic flux. Here, we don't have a closed loop, so we need to find the emf generated along the length of the rod.
03

Calculate Magnetic Flux

The magnetic flux through the rod is given by \( \Phi_B = B \times A \), where \(B\) is the magnetic field strength and \(A\) is the area swept by the rod. For a thin rotating rod, \(A\) at any position is \(r \times v \), with \(v = \omega r\) and \(r\) ranging from 0 to \(\ell\).
04

Integrate emf Across the Rod's Length

To find the total emf, integrate from 0 to \(\ell\): \[ \text{emf} = \int_{0}^{\ell} (B \cdot \omega \cdot r) \, dr \].
05

Solve the Integral

Execute the integral: \[ \text{emf} = B \cdot \omega \cdot \int_{0}^{\ell} r \, dr = B \cdot \omega \cdot \left[ \frac{r^2}{2} \right]_{0}^{\ell} = B \cdot \omega \cdot \frac{\ell^2}{2} \].
06

Final Step: Conclusion

The emf developed between the ends of the rod due to its rotation in the magnetic field is \( \frac{1}{2} B \omega \ell^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law of Induction
At the heart of the problem is Faraday's Law of Induction, a fundamental principle that helps us understand how electromotive force (emf) is generated. This law tells us that an emf is induced whenever a conductor experiences a change in magnetic flux. Now, you may wonder what magnetic flux is—more on that later.

To compute the induced emf, Faraday’s law provides us with the equation: \[ \text{emf} = -\frac{d\Phi_B}{dt} \]The symbol \( \Phi_B \) stands for the magnetic flux, which refers to the quantity of the magnetic field passing through a loop or a surface. In this problem, although we don't have a looped circuit, the concept still applies to the entire length of the rod.

Key points to remember about Faraday's Law are:
  • The faster the change in magnetic flux, the greater the induced emf.
  • Induced emf acts to oppose the change in flux, a phenomenon known as Lenz’s Law.
  • Though the formula has a negative sign, what's crucial is the magnitude of the induced emf—which is determined by the integration over the rod's length in this scenario.
Magnetic Flux
To understand how emf is generated, we need to dive into the concept of magnetic flux. Think of magnetic flux as the number of magnetic field lines passing through a surface or area. It's like water flowing through a net—the more significant the area or stronger the magnetic field, the greater the flux.

Mathematically, magnetic flux \( \Phi_B \) is calculated using the formula:\[ \Phi_B = B \times A \]where \( B \) is the magnetic field strength, and \( A \) is the area that the magnetic field lines penetrate.

In the case of the rotating rod, this area is dynamic. As the rod spins, it sweeps out a sector that changes with time. It's like the way a clock's hand moves, covering a wedge-shaped area over time.

As the rod rotates in the magnetic field, the magnetic flux through the area it sweeps varies, allowing the calculation of emf using the relationship:
  • Larger magnetic fields, \( B \), result in more magnetic flux.
  • Larger swept areas \( A \) indicate more lines of the magnetic field intersecting with the rod.
Rotating Rod
In this exercise, we analyze a rotating rod within a magnetic field, which is a fantastic application of electromagnetic principles. Imagine a thin metal rod, pivoted at one end, moving in a circle—much like a second hand on a clock but rotating far more quickly.

As it spins with an angular velocity \( \omega \), every point on the rod travels at a different linear speed. Points closer to the pivot move slower than those on the outer edge. The linear velocity \( v \) of a point from the pivot is \( v = \omega \times r \), where \( r \) is the distance from the pivot.

Why does this matter? Because the motion through the magnetic field induces an emf. The faster the motion at a given point, the higher the potential difference generated at that point. When the rod rotates:
  • The section farthest from the pivot contributes the most to the emf because it travels the fastest.
  • The total emf between the rod's ends equals integrating contributions from all points along the rod.
Using calculus here reveals that the total emf depends on the square of the length of the rod \( \ell \). Thus, the longer the rod, the more significant the induced potential difference.

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Most popular questions from this chapter

(I) Neon signs require \(12 \mathrm{kV}\) for their operation. To operate from a \(240-\mathrm{V}\) line, what must be the ratio of secondary to primary turns of the transformer? What would the voltage output be if the transformer were connected backward?

A search coil for measuring \(B\) (also called a flip coil) is a small coil with \(N\) turns, each of cross-sectional area \(A .\) It is connected to a so-called ballistic galvanometer, which is a device to measure the total charge \(Q\) that passes through it in a short time. The flip coil is placed in the magnetic field to be measured with its face perpendicular to the field. It is then quickly rotated \(180^{\circ}\) about a diameter. Show that the total charge \(Q\) that flows in the induced current during this short "flip" time is proportional to the magnetic field \(B\). In particular, show that \(B\) is given by $$ B=\frac{Q R}{2 N A} $$ where \(R\) is the total resistance of the circuit, including that of the coil and that of the ballistic galvanometer which measures the charge \(Q\).

A coil with 150 turns, a radius of \(5.0 \mathrm{~cm},\) and a resistance of \(12 \Omega\) surrounds a solenoid with 230 turns \(/ \mathrm{cm}\) and a radius of \(4.5 \mathrm{~cm}\); see Fig. \(29-50 .\) The current in the solenoid changes at a constant rate from 0 to \(2.0 \mathrm{~A}\) in \(0.10 \mathrm{~s}\). Calculate the magnitude and direction of the induced current in the 150 -turn coil.

Two resistanceless rails rest 32 \(\mathrm{cm}\) apart on a \(6.0^{\circ}\) ramp. They are joined at the bottom by a \(0.60-\Omega\) resistor. At the top a copper bar of mass 0.040 \(\mathrm{kg}\) (ignore its resistance) is laid across the rails. The whole apparatus is immersed in a vertical 0.55 - field. What is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails?

(II) The magnetic flux through each loop of a 75 -loop coil is given by \(\left(8.8 t-0.51 t^{3}\right) \times 10^{-2} \mathrm{~T} \cdot \mathrm{m}^{2},\) where the time \(t\) is in seconds. (a) Determine the emf \(\mathscr{E}\) as a function of time. (b) What is \(\mathscr{E}\) at \(t=1.0 \mathrm{~s}\) and \(t=4.0 \mathrm{~s} ?\)

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