The rotating-cylinder viscometer in the image shears the fluid in a narrow clearance Ar as shown. Assume a linear velocity distribution in the gaps. - ΔΩ Viscous fluid Solid cylinder →Ar<
The rotating-cylinder viscometer in the image shears the fluid in a narrow clearance Ar as shown. Assume a linear velocity distribution in the gaps. - ΔΩ Viscous fluid Solid cylinder →Ar<
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
100%
![## Understanding Magnetic Moment (μ) Equations
This section features multiple choice options to determine the correct formula for the magnetic moment (μ). Each option is associated with a radio button that allows the selection of one answer.
1. **Option 1:**
\[
\mu = \frac{2 \pi N M A R}{\Omega R^2 L^3}
\]
- This formula expresses the magnetic moment (μ) as a function of various parameters: the number of turns (N), magnetization (M), cross-sectional area (A), radius (R), angular velocity (Ω), and length (L).
2. **Option 2:**
\[
\mu = \frac{2 \pi N M A R}{\Omega R^2 L}
\]
- In this equation, the magnetic moment (μ) depends on the variables N, M, A, R, Ω, and L. Similar to the first option but with different denominators.
3. **Option 3:**
\[
\mu = \frac{M A R}{2 \pi \Omega R^2 L}
\]
- Here, the magnetic moment (μ) is presented with a different coefficient and slightly altered form involving M, A, R, Ω, and L.
4. **Option 4:**
\[
\mu = \frac{M A R}{2 \pi \Omega R L}
\]
- This formula again shows the magnetic moment (μ) related to M, A, R, Ω, and L, with yet another variation in terms.
When choosing the correct formula, ensure to consider the correct physical constants and dimensional consistency for your specific application. Select the appropriate circle based on the derived formula or given experimental data. Each option is designed to test your understanding of the relationship between magnetic moment and other physical quantities.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbab71d29-f3d9-43ce-83f4-a263357daa0d%2F4ebe890d-2177-44ed-9362-346a6bb7678c%2Fybe4ze0p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Understanding Magnetic Moment (μ) Equations
This section features multiple choice options to determine the correct formula for the magnetic moment (μ). Each option is associated with a radio button that allows the selection of one answer.
1. **Option 1:**
\[
\mu = \frac{2 \pi N M A R}{\Omega R^2 L^3}
\]
- This formula expresses the magnetic moment (μ) as a function of various parameters: the number of turns (N), magnetization (M), cross-sectional area (A), radius (R), angular velocity (Ω), and length (L).
2. **Option 2:**
\[
\mu = \frac{2 \pi N M A R}{\Omega R^2 L}
\]
- In this equation, the magnetic moment (μ) depends on the variables N, M, A, R, Ω, and L. Similar to the first option but with different denominators.
3. **Option 3:**
\[
\mu = \frac{M A R}{2 \pi \Omega R^2 L}
\]
- Here, the magnetic moment (μ) is presented with a different coefficient and slightly altered form involving M, A, R, Ω, and L.
4. **Option 4:**
\[
\mu = \frac{M A R}{2 \pi \Omega R L}
\]
- This formula again shows the magnetic moment (μ) related to M, A, R, Ω, and L, with yet another variation in terms.
When choosing the correct formula, ensure to consider the correct physical constants and dimensional consistency for your specific application. Select the appropriate circle based on the derived formula or given experimental data. Each option is designed to test your understanding of the relationship between magnetic moment and other physical quantities.

* **Solid cylinder**: Represented by the rectangular block in the middle.
* **Viscous fluid** \( \mu \): The fluid surrounding the cylinder.
* **Rotation**: Indicated by the symbol \( \Omega \) on top of the cylinder.
* **Dimensions**:
* \( R \) – Radius of the solid cylinder.
* \( L \) – Length of the solid cylinder.
* \( \Delta r \) – Narrow clearance gap between the cylinder and the container wall.
#### Problem Statement
If the driving torque \( M \) is measured, find an expression for \( \mu \) by neglecting the bottom friction.
---
### Solution Approach
To solve this, consider the balance of forces and the shear stress in the fluid film. Utilizing the torque \( M \) and the assumption of linear velocity distribution, derive the relationship between torque, fluid viscosity, and the geometric parameters of the system. Detailed steps may include:
1. **Shear Stress Calculation**:
\[
\tau = \mu \left( \frac{dV}{dy} \right)
\]
where \( V \) is the velocity and \( y \) is the radial position.
2. **Velocity Gradient**:
\[
\frac{dV}{dy} \approx \frac{\Omega R}{\Delta r}
\]
3. **Torque and Shear Stress Relationship**:
Considering torque balance and integrating over the cylinder surface:
\[
M = 2 \pi R^2 L \tau
\]
4. **Expressing Viscosity \( \mu \)**:
Substituting the expressions for velocity gradient and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbab71d29-f3d9-43ce-83f4-a263357daa0d%2F4ebe890d-2177-44ed-9362-346a6bb7678c%2Fhnmcygt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding the Rotating-Cylinder Viscometer
#### Required Information
The rotating-cylinder viscometer in the image shears the fluid in a narrow clearance \( \Delta r \) as shown. Assume a linear velocity distribution in the gaps.
#### Diagram Description
The diagram illustrates a solid cylinder of radius \( R \) immersed in a viscous fluid with viscosity \( \mu \). The cylinder has a height \( L \) and rotates with an angular velocity \( \Omega \). The clearance between the cylinder and the container wall is \( \Delta r \), which is much smaller than \( R \) (\( \Delta r \ll R \)).

* **Solid cylinder**: Represented by the rectangular block in the middle.
* **Viscous fluid** \( \mu \): The fluid surrounding the cylinder.
* **Rotation**: Indicated by the symbol \( \Omega \) on top of the cylinder.
* **Dimensions**:
* \( R \) – Radius of the solid cylinder.
* \( L \) – Length of the solid cylinder.
* \( \Delta r \) – Narrow clearance gap between the cylinder and the container wall.
#### Problem Statement
If the driving torque \( M \) is measured, find an expression for \( \mu \) by neglecting the bottom friction.
---
### Solution Approach
To solve this, consider the balance of forces and the shear stress in the fluid film. Utilizing the torque \( M \) and the assumption of linear velocity distribution, derive the relationship between torque, fluid viscosity, and the geometric parameters of the system. Detailed steps may include:
1. **Shear Stress Calculation**:
\[
\tau = \mu \left( \frac{dV}{dy} \right)
\]
where \( V \) is the velocity and \( y \) is the radial position.
2. **Velocity Gradient**:
\[
\frac{dV}{dy} \approx \frac{\Omega R}{\Delta r}
\]
3. **Torque and Shear Stress Relationship**:
Considering torque balance and integrating over the cylinder surface:
\[
M = 2 \pi R^2 L \tau
\]
4. **Expressing Viscosity \( \mu \)**:
Substituting the expressions for velocity gradient and
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images

Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you

Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press

Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON

Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education

Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press

Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON

Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education

Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY

Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning

Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY