### Description of the Geometric Shape and Problem StatementConsider the special shape pictured in the diagram. It is a cylinder, centered on the origin with its axis oriented along the z-axis. The cylinder is partially hollowed with cone-shaped cavities at the top and bottom.### Diagram Explanation- **Left Diagram:** Displays a solid cylinder with a pink color to show the material.- **Middle Diagram:** Illustrates the internal cone-shaped cavities from a vertical perspective, where the hollow regions are represented through shading or hatching.- **Right Diagram:** Shows a cross-sectional side view of the cylinder. - The radius is labeled as **\( a \)**. - The height is labeled as **\( 2a \)**. - Shaded regions show the solid parts of the object.### Problem Specifications- **Radius of the Object:** \( a \)- **Height of the Object:** \( 2a \)- **Solid Part's Volume Charge Density:** Uniform, denoted as \( \rho_0 \)- Assume that the object spins counterclockwise about its cylinder axis with an angular frequency \( \omega \).### Problem QueryThe task is to determine which operations are part of calculating the magnitude of the current density associated with the motion of the rotating object as a function of \( r \). Select all applicable operations. - differentiating the charge density with respect to time- integrating over r with 0 and a as the limits of integration- integrating over the azimuthal angle φ with 0 and 2π as the limits of integration- expressing (in terms of r) the cross-sectional area of the part of the spinning object that is bounded by the region where (x^2+y^2) ≤ r^2

In this question we have to answer what operations were used while determining the magnitude of the current density.Solution:Consider the expression for charge density, given as:ρ = dqdvRe-arranging the eqaution, we have:dq = ρdvAnd the expression for current density is given as:J =didA=dqdt·dA , where, q is charge i is the current and, A is cross-sectional areaApplying the value of dq in above equation, we get:J = ρdvdt·dA - 1Now, consider the change in length of current carrying conductor, given as:dl = r dθwhere, r is radiusThus, we have the relation:dV = dl·dA = r dθ dAApplying the same in equation 1, we have current density equal to:J = ρr dθ dAdt·dA , where dθdt=ω∴ J = ρrω where, ρ is surface charge density.Therefore, the correct answer will be 'differentiating the charge density with respect to time'

Science

Physics

Consider the special shape pictured in the diagram below. It is a cylinder, centered on the origin with its axis oriented along z, and it has been partially hollowed to leave two cone-shaped cavities at the top and bottom of the cylinder.

Consider the special shape pictured in the diagram below. It is a cylinder, centered on the origin with its axis oriented along z, and it has been partially hollowed to leave two cone-shaped cavities at the top and bottom of the cylinder.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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