/*! 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 59 A slender rod, 0.240 m long, rot... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 59

A slender rod, 0.240 m long, rotates with an angular speed of 8.80 rad/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 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 rad/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?

Short Answer

Expert verified
(a) 0.164 V; (b) 0.164 V; (c) 0.082 V; 0.041 V between center and end.

Step by step solution

01

Understanding the Problem

A slender rod 0.240 m long is rotating with an angular speed of 8.80 rad/s in a uniform magnetic field of 0.650 T. We're asked to calculate the induced emf and the potential differences under two different conditions: rotation around one end, and rotation about the center.
02

Calculate Induced EMF for Rotation about One End

The formula for the induced emf in a rotating rod is given by: \( \text{emf} = \frac{1}{2} B \omega L^2 \). Here, \( B = 0.650 \, T \), \( \omega = 8.80 \, \text{rad/s} \), and \( L = 0.240 \, m \). Substitute these values to get: \[ \text{emf} = \frac{1}{2} \times 0.650 \times 8.80 \times (0.240)^2 = 0.164 \text{ volts}. \]
03

Determine Potential Difference Between Ends for Rotation about One End

The potential difference between the ends is equal to the induced emf calculated in Step 1, which is 0.164 V.
04

Calculate Potential Difference for Rotation about Center

For this scenario, the effective length of the rod is reduced as the rod rotates about its center. The induced emf is given by \( \text{emf} = \frac{1}{2} B \omega (\frac{L}{2})^2 \times 2 \), as there are two halves of the rod. Substitute the values: \[ \text{emf} = \frac{1}{2} \times 0.650 \times 8.80 \times (0.240/2)^2 \times 2 = 0.082 \text{ volts}. \]
05

Determine Potential Difference Between Center and End

The potential difference from the center to one end is half of the total induced emf for rotation about the center, which is \( \frac{0.082}{2} = 0.041 \, \text{V} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Magnetic Field
A magnetic field is an invisible field that exerts magnetic forces on moving electric charges and magnetic dipoles. It is a vital concept in understanding how electromotive force (emf) is induced in moving conductors. In this exercise, a uniform magnetic field with a magnitude of 0.650 Teslas (T) is present. This magnetic field is perpendicular to the plane of rotation of the rod, influencing the electrons inside the rod to move and create an emf.

Key characteristics of a magnetic field:
  • It is represented by the symbol 'B'.
  • The strength of the magnetic field, or its magnitude, influences the amount of induced emf.
  • Uniformity means the field's strength and direction are constant across the area of interest.
It is crucial to understand that the magnetic field is the driving factor behind the inducement of emf when a conductor, like our rod, moves or rotates within it.
Rotation about an Axis
Rotation about an axis refers to the motion where an object spins around an imaginary line called the axis of rotation. In our exercise, the rod rotates with an angular speed around two different axes: once through one end, and once through its center.

Important aspects of rotation:
  • The axis of rotation defines the path of rotation. Changing the axis alters the distribution of mass and affects physical quantities like induced emf.
  • For a rod rotating about one end, the entire length contributes to the generation of emf, whereas for rotation about the center, each half contributes separately.
Understanding the concept of rotation is essential for calculating how the induced emf changes based on the axis of rotation.
Potential Difference
Potential difference, often referred to as voltage, is the difference in electric potential energy per unit charge between two points in a circuit. In the exercise, you are dealing with the potential difference between various parts of the rod, which results from the induced emf.

Key points about potential difference:
  • It is calculated from the induced emf using the relation that the emf equals the potential difference between the two ends of the rod.
  • For rotation about one end, the potential difference equals 0.164 V because the entire rod is utilized.
  • For rotation about the center, potential differences differ: across the whole rod, it's 0.082 V, while from center to one end, it's 0.041 V.
Grasping the concept of potential difference helps to understand the effects of rotation on the electrical characteristics within the rod.
Angular Speed
Angular speed, denoted by the symbol \( \omega \), quantifies how quickly an object rotates around its axis. It is given in units like radians per second (rad/s). In this exercise, the rod rotates at an angular speed of 8.80 rad/s.

Why angular speed matters:
  • Angular speed is directly proportional to the induced emf, meaning that higher speeds result in larger emf.
  • It is a critical component in the formula \( \text{emf} = \frac{1}{2} B \omega L^2 \) for calculating the emf when rotating about one end.
  • The consistency of angular speed ensures that calculations made for potential differences and emf remain valid.
Understanding angular speed's role is essential to predict how different velocities of rotation will influence the effectiveness of emf induction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The magnetic field within a long, straight solenoid with a circular cross section and radius \(R\) is increasing at a rate of \(dB/dt\). (a) What is the rate of change of flux through a circle with radius \(r_1\) inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance \(r_1\) from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field \(outside\) the solenoid, at a distance \(r_2\) from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance \(r\) from the axis from \(r =\) 0 to \(r = 2R\). (e) What is the magnitude of the induced emf in a circular turn of radius R/2 that has its center on the solenoid axis? (f) What is the magnitude of the induced emf if the radius in part (e) is \(R\)? (g) What is the induced emf if the radius in part (e) is 2\(R\)?

In a region of space, a magnetic field points in the +\(x\)-direction (toward the right). Its magnitude varies with position according to the formula \(B_x = B_0 + bx\), where \(B_0\) and \(b\) are positive constants, for \(x \geq\) 0. A flat coil of area \(A\) moves with uniform speed \(v\) from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin, what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answers to parts (a) and (b)?

A circular conducting ring with radius \(r_0 =\) 0.0420 m lies in the xy-plane in a region of uniform magnetic field \(\overrightarrow{B} = B_0 [1 - 3(t/t_0)^2 + 2(t/t_0)^3]\hat{k}\). In this expression, \(t_0 =\) 0.0100 s and is constant, \(t\) is time, \(\hat{k}\) is the unit vector in the +\(z\)-direction, and \(B_0\) = 0.0800 T and is constant. At points \(a\) and \(b\) (Fig. P29.58) there is a small gap in the ring with wires leading to an external circuit of resistance \(R =\) 12.0 \(\Omega\). There is no magnetic field at the location of the external circuit. (a) Derive an expression, as a function of time, for the total magnetic flux \(\Phi_B\) through the ring. (b) Determine the emf induced in the ring at time \(t =\) 5.00 \(\times\) 10\(^{-3}\) s. What is the polarity of the emf? (c) Because of the internal resistance of the ring, the current through \(R\) at the time given in part (b) is only 3.00 mA. Determine the internal resistance of the ring. (d) Determine the emf in the ring at a time \(t =\) 1.21 \(\times\) 10\(^{-2}\) s. What is the polarity of the emf? (e) Determine the time at which the current through \(R\) reverses its direction.

A very long, rectangular loop of wire can slide without friction on a horizontal surface. Initially the loop has part of its area in a region of uniform magnetic field that has magnitude \(B =\) 2.90 T and is perpendicular to the plane of the loop. The loop has dimensions 4.00 cm by 60.0 cm, mass 24.0 g, and resistance \(R =\) 5.00 \(\times\) 10\(^{-3} \Omega\). The loop is initially at rest; then a constant force \(F_{ext}\) = 0.180 N is applied to the loop to pull it out of the field (Fig. P29.46). (a) What is the acceleration of the loop when \(v =\) 3.00 cm/s? (b) What are the loop's terminal speed and acceleration when the loop is moving at that terminal speed? (c) What is the acceleration of the loop when it is completely out of the magnetic field?

An airplane propeller of total length \(L\) rotates around its center with angular speed \(\omega\) in a magnetic field that is perpendicular to the plane of rotation. Modeling the propeller as a thin, uniform bar, find the potential difference between (a) the center and either end of the propeller and (b) the two ends. (c) If the field is the earth's field of 0.50 G and the propeller turns at 220 rpm and is 2.0 m long, what is the potential difference between the middle and either end? It this large enough to be concerned about?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Measurements

Read Explanation

Astrophysics

Read Explanation

Force

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.