Question

2: Poynting energy flow around a battery Panels (a) and (b) below illustrate two perspectives of a cylindrical, ideal (resistance-less) battery of length L and radius R. The battery is connected to an external circuit (not illustrated) and operates under static conditions, meaning E and Jf are steady. You may assume the battery is composed of a homogeneous Ohmic material, such that E and Jf are everywhere constant within the battery. ^(2) (a) Evaluate the magnitude of H along the pink Amperean loop in panel (a), using the Ampere-Maxwell law for the free current. The pink loop has radius R but is contained just within the battery, meaning it doesn't encircle any surface charge. ^(3) You should not assume H = B/μ0 within the battery, i.e., you should allow the battery to have a nontrivial magnetization. (b) Evaluate the energy flux through the pink sheath illustrated in panel (c), i.e., calculate the area integral of the Poynting vector: ∫ E × H * da. This sheath is cylindrical with radius R, is uncapped on both ends, ^(4) and is contained just within the battery; assume da points outward. Express your final answer in terms of the free current If = |Jf|πR^2 and the potential drop across the external circuit V = |E|L. (c) Is Poynting's energy current flowing in or out of the battery? If 'in', where is the energy coming from? If 'out', where is the energy going? Explain in one sentence. The battery is immersed in vacuum. The surface charge density of the battery is inhomogeneous and illustrated by the green plus and minus signs in panel (b); the distribution of the charge density is identical to what one would expect of an Ohmic conductor, which we determined in a previous poll question. Our goal is to determine the Poynting vector E × H just outside the battery. We'll do this in a few steps: (d) By considering the line integral of E along the orange rectangular loop (of infinitesimal thickness) in panel (b), apply Faraday's law to determine if the surface-parallel components of E are continuous or discontinuous across the surface. (e) Read the Gaussian-pillbox argument for D-D= in 7.3.6 of Griffiths; σf is the surface charge density and should not be confused with a conductivity. Using this equation, determine the perpendicular component E(r) of the electric field at a point (in vacuum) just outside the battery, e.g., at representative points A or B in panel (b). Express E(r) in terms of the surface charge density σf(r) and the vacuum permittivity. (f) Read the boundary condition for the surface-parallel component of H in the same section of Griffiths. Assuming that Kf=0, evaluate the magnitude of H along the purple Amperean loop in panel (a). The purple loop has radius R but lies just outside the battery, meaning it encircles the surface charge. ^(5)

Name: 2 poynting energy flow around a battery panels a and b below illustrate two perspectives of a cylindrical ideal resistance less battery of length l and radius r the battery is connected to a 52939
Uploaded: 2022-07-05T05:43:21-08:00
Duration: 2 min 32 s
Channel: Suman K
Description: 2 poynting energy flow around a battery panels a and b below illustrate two perspectives of a cylindrical ideal resistance less battery of length l and radius r the battery is connected to a 52939

          2: Poynting energy flow around a battery
Panels (a) and (b) below illustrate two perspectives of a cylindrical, ideal (resistance-less) battery of length L and radius R.
The battery is connected to an external circuit (not illustrated) and operates under static conditions, meaning E and Jf are steady.
You may assume the battery is composed of a homogeneous Ohmic material, such that E and Jf are everywhere constant within the battery. ^(2)
(a) Evaluate the magnitude of H along the pink Amperean loop in panel (a), using the Ampere-Maxwell law for the free current.
The pink loop has radius R but is contained just within the battery, meaning it doesn't encircle any surface charge. ^(3) You should
not assume H = B/μ0 within the battery, i.e., you should allow the battery to have a nontrivial magnetization.
(b) Evaluate the energy flux through the pink sheath illustrated in panel (c), i.e., calculate the area integral of the Poynting vector:
∫ E × H * da. This sheath is cylindrical with radius R, is uncapped on both ends, ^(4) and is contained just within the battery;
assume da points outward. Express your final answer in terms of the free current If = |Jf|πR^2 and the potential drop across
the external circuit V = |E|L.
(c) Is Poynting's energy current flowing in or out of the battery? If 'in', where is the energy coming from? If 'out', where is the
energy going? Explain in one sentence.
The battery is immersed in vacuum. The surface charge density of the battery is inhomogeneous and illustrated by the green plus
and minus signs in panel (b); the distribution of the charge density is identical to what one would expect of an Ohmic conductor,
which we determined in a previous poll question. Our goal is to determine the Poynting vector E × H just outside the battery.
We'll do this in a few steps:
(d) By considering the line integral of E along the orange rectangular loop (of infinitesimal thickness) in panel (b), apply Faraday's law to determine if the surface-parallel components of E are continuous or discontinuous across the surface.
(e) Read the Gaussian-pillbox argument for D-D= in 7.3.6 of Griffiths; σf is the surface charge density and should
not be confused with a conductivity. Using this equation, determine the perpendicular component E(r) of the electric field at a
point (in vacuum) just outside the battery, e.g., at representative points A or B in panel (b). Express E(r) in terms of the surface
charge density σf(r) and the vacuum permittivity.
(f) Read the boundary condition for the surface-parallel component of H in the same section of Griffiths. Assuming that Kf=0,
evaluate the magnitude of H along the purple Amperean loop in panel (a). The purple loop has radius R but lies just outside the
battery, meaning it encircles the surface charge. ^(5)

2 poynting energy flow around a battery panels a and b below illustrate two perspectives of a cylindrical ideal resistance less battery of length l and radius r the battery is connected to a 52939

Added by David K.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Suman K

07/05/2022

Step 1

This law states ∮H⋅dl = I_f, where I_f is the free current passing through the loop. Since the loop is contained just within the battery and doesn't encircle any surface charge, we need to consider the magnetization of the battery. Show more…

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2: Poynting energy flow around a battery Panels (a) and (b) below illustrate two perspectives of a cylindrical, ideal (resistance-less) battery of length L and radius R. The battery is connected to an external circuit (not illustrated) and operates under static conditions, meaning E and Jf are steady. You may assume the battery is composed of a homogeneous Ohmic material, such that E and Jf are everywhere constant within the battery. ^(2) (a) Evaluate the magnitude of H along the pink Amperean loop in panel (a), using the Ampere-Maxwell law for the free current. The pink loop has radius R but is contained just within the battery, meaning it doesn't encircle any surface charge. ^(3) You should not assume H = B/μ0 within the battery, i.e., you should allow the battery to have a nontrivial magnetization. (b) Evaluate the energy flux through the pink sheath illustrated in panel (c), i.e., calculate the area integral of the Poynting vector: ∫ E × H * da. This sheath is cylindrical with radius R, is uncapped on both ends, ^(4) and is contained just within the battery; assume da points outward. Express your final answer in terms of the free current If = |Jf|πR^2 and the potential drop across the external circuit V = |E|L. (c) Is Poynting's energy current flowing in or out of the battery? If 'in', where is the energy coming from? If 'out', where is the energy going? Explain in one sentence. The battery is immersed in vacuum. The surface charge density of the battery is inhomogeneous and illustrated by the green plus and minus signs in panel (b); the distribution of the charge density is identical to what one would expect of an Ohmic conductor, which we determined in a previous poll question. Our goal is to determine the Poynting vector E × H just outside the battery. We'll do this in a few steps: (d) By considering the line integral of E along the orange rectangular loop (of infinitesimal thickness) in panel (b), apply Faraday's law to determine if the surface-parallel components of E are continuous or discontinuous across the surface. (e) Read the Gaussian-pillbox argument for D-D= in 7.3.6 of Griffiths; σf is the surface charge density and should not be confused with a conductivity. Using this equation, determine the perpendicular component E(r) of the electric field at a point (in vacuum) just outside the battery, e.g., at representative points A or B in panel (b). Express E(r) in terms of the surface charge density σf(r) and the vacuum permittivity. (f) Read the boundary condition for the surface-parallel component of H in the same section of Griffiths. Assuming that Kf=0, evaluate the magnitude of H along the purple Amperean loop in panel (a). The purple loop has radius R but lies just outside the battery, meaning it encircles the surface charge. ^(5)