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A cylindrical beam, length L and radius R, of particles, each particle with positive charge q, moves with speed v to create an axial current (flow parallel to the axis) such as that in example 26.8 in the text. The current, I, is related to the speed via the relationship I = puA, where p is the charge per unit volume and A is the cross sectional area of the beam. The charge density p is uniform. - As a result of the current and the charges in the beam, each particle sees and responds to a magnetic field created at its location by the rest of the current and an electric field due to the rest of the charge. We will treat the beam as an infinitely long, straight axial current so that we can use Ampere's law. (a) Consider first the directions of the electric and magnetic forces (either identify the directions of the fields and their resulting forces, or consider the fact that we have identical charges and moving in the same direction, each making a parallel current.) What are these directions? Now let's find a mathematical expression for the magnitude and direction of the net force on each charge and the speed at which these are equal: (b) Find B at r < R, using the results of example 26.8 for test points inside the current (r < R) and from that the force on the charge at r. See the guide for hints. (c) Use Gauss's law (you might want to write it in terms of the charge density p = Q/(TR²L) rather than the total charge, Q, in the cylinder) to find the electric field at r < R and, from that, the electric force on the charge at r.

A cylindrical beam, length L and radius R, of particles, each particle with positive charge q, moves with speed v to create an axial current (flow parallel to the axis) such as that in example 26.8 in the text. The current, I, is related to the speed via the relationship I = puA, where p is the charge per unit volume and A is the cross sectional area of the beam. The charge density p is uniform. - As a result of the current and the charges in the beam, each particle sees and responds to a magnetic field created at its location by the rest of the current and an electric field due to the rest of the charge. We will treat the beam as an infinitely long, straight axial current so that we can use Ampere's law. (a) Consider first the directions of the electric and magnetic forces (either identify the directions of the fields and their resulting forces, or consider the fact that we have identical charges and moving in the same direction, each making a parallel current.) What are these directions? Now let's find a mathematical expression for the magnitude and direction of the net force on each charge and the speed at which these are equal: (b) Find B at r < R, using the results of example 26.8 for test points inside the current (r < R) and from that the force on the charge at r. See the guide for hints. (c) Use Gauss's law (you might want to write it in terms of the charge density p = Q/(TR²L) rather than the total charge, Q, in the cylinder) to find the electric field at r < R and, from that, the electric force on the charge at r.

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Answer referring to example 26.7
EXAMPLE 26.7 Using Ampère's Law: Outside and Inside a Wire Worked Example with Variation Problems A long, straight wire of radius R carries a current I distributed uni- formly over its cross section. Find the magnetic field (a) outside and (b) inside the wire. INTERPRET We follow our strategy and identify this as a situation with line symmetry. Therefore, the field depends only on the radial distance from the wire's central axis. DEVELOP Magnetic field lines encircle their source, so the only field lines consistent with the symmetry are concentric circles. We've sketched some of these circular field lines in Fig. 26.31. The field is tangent to the field lines and, by symmetry, has the same magnitude B everywhere on a field line. So the field lines themselves make good Ampèrian loops. EVALUATE The field is everywhere parallel to a circular Ampèrian loop, and its magnitude is constant on the loop, so for a loop of radius r, B.di becomes B fdl, or just 27rB, because fdl is the circumference 27r. This is true both outside and inside the wire. ● FIGURE 26.31 Cross section of a long cylindrical wire. Any field line can serve as an Ampèrian loop. Inside the wire, the loop's radius r is less than the wire's radius R; outside, r> R. B We'll first answer (a): For any loop outside the wire, the encircled current is the total rent I. B-dl to o times the encircled Equating our expression for current gives 27rB = μol, so B = Hol 2πr field outside any current distribution with line symmetry (26.18) (continued)
Transcribed Image Text:EXAMPLE 26.7 Using Ampère's Law: Outside and Inside a Wire Worked Example with Variation Problems A long, straight wire of radius R carries a current I distributed uni- formly over its cross section. Find the magnetic field (a) outside and (b) inside the wire. INTERPRET We follow our strategy and identify this as a situation with line symmetry. Therefore, the field depends only on the radial distance from the wire's central axis. DEVELOP Magnetic field lines encircle their source, so the only field lines consistent with the symmetry are concentric circles. We've sketched some of these circular field lines in Fig. 26.31. The field is tangent to the field lines and, by symmetry, has the same magnitude B everywhere on a field line. So the field lines themselves make good Ampèrian loops. EVALUATE The field is everywhere parallel to a circular Ampèrian loop, and its magnitude is constant on the loop, so for a loop of radius r, B.di becomes B fdl, or just 27rB, because fdl is the circumference 27r. This is true both outside and inside the wire. ● FIGURE 26.31 Cross section of a long cylindrical wire. Any field line can serve as an Ampèrian loop. Inside the wire, the loop's radius r is less than the wire's radius R; outside, r> R. B We'll first answer (a): For any loop outside the wire, the encircled current is the total rent I. B-dl to o times the encircled Equating our expression for current gives 27rB = μol, so B = Hol 2πr field outside any current distribution with line symmetry (26.18) (continued)
A cylindrical beam, length L and radius R, of particles, each particle with positive charge q, moves with speed v to create an axial current (flow parallel to the axis) such as that in example 26.8 in the text. The current, I, is related to the speed via the relationship I = pu A, where P is the charge per unit volume and A is the cross sectional area of the beam. The charge density p is uniform. As a result of the current and the charges in the beam, each particle sees and responds to a magnetic field created at its location by the rest of the current and an electric field due to the rest of the charge. We will treat the beam as an infinitely long, straight axial current so that we can use Ampere's law. (a) Consider first the directions of the electric and magnetic forces (either identify the directions of the fields and their resulting forces, or consider the fact that we have identical charges and moving in the same direction, each making a parallel current.) What are these directions? Now let's find a mathematical expression for the magnitude and direction of the net force on each charge and the speed at which these are equal: (b) Find B at r < R, using the results of example 26.8 for test points inside the current (r< R) and from that the force on the charge at r. See the guide for hints. (c) Use Gauss's law (you might want to write it in terms of the charge density p= Q/(R²L) rather than the total charge, Q, in the cylinder) to find the electric field at r < R and, from that, the electric force on the charge at r. (d) Finally, determine the speed at which the two forces are equal. Simplify your results using the relationship between current and the flow of charge, I = puA and the fact that the speed of light is c=1/√√600. When the speed is less than this, what is the net effect on the beam? C
Transcribed Image Text:A cylindrical beam, length L and radius R, of particles, each particle with positive charge q, moves with speed v to create an axial current (flow parallel to the axis) such as that in example 26.8 in the text. The current, I, is related to the speed via the relationship I = pu A, where P is the charge per unit volume and A is the cross sectional area of the beam. The charge density p is uniform. As a result of the current and the charges in the beam, each particle sees and responds to a magnetic field created at its location by the rest of the current and an electric field due to the rest of the charge. We will treat the beam as an infinitely long, straight axial current so that we can use Ampere's law. (a) Consider first the directions of the electric and magnetic forces (either identify the directions of the fields and their resulting forces, or consider the fact that we have identical charges and moving in the same direction, each making a parallel current.) What are these directions? Now let's find a mathematical expression for the magnitude and direction of the net force on each charge and the speed at which these are equal: (b) Find B at r < R, using the results of example 26.8 for test points inside the current (r< R) and from that the force on the charge at r. See the guide for hints. (c) Use Gauss's law (you might want to write it in terms of the charge density p= Q/(R²L) rather than the total charge, Q, in the cylinder) to find the electric field at r < R and, from that, the electric force on the charge at r. (d) Finally, determine the speed at which the two forces are equal. Simplify your results using the relationship between current and the flow of charge, I = puA and the fact that the speed of light is c=1/√√600. When the speed is less than this, what is the net effect on the beam? C
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