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A charge of 4.0 μC is distributed uniformly around a thin ring of insulating material. The ring has a radius of 0.20 m and rotates at 2.0 × 10 4 rev/min around the axis that passes through its center and is perpendicular to the plane of the ring. What is the magnetic field at the center of the ring?

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A thin, nonconducting disk of radius R is free to rotate around the axis that passes through its center and is perpendicular to the face of the disk. The disk is charged uniformly with a total charge q . If the disk rotates at a constant angular velocity ω , what is the magnetic field at its center?

B = μ 0 σ ω 2 R

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Consider the disk in the previous problem. Calculate the magnetic field at a point on its central axis that is a distance y above the disk.

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Consider the axial magnetic field B v = μ 0 I R 2 / 2 ( y 2 + R 2 ) 3 / 2 of the circular current loop shown below. (a) Evaluate a a B y d y . Also show that lim a a a B y d y = μ 0 I . (b) Can you deduce this limit without evaluating the integral? ( Hint: See the accompanying figure.)

This picture shows the circular current loop I with the magnetic field B perpendicular to the plane of the loop.

derivation

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The current density in the long, cylindrical wire shown in the accompanying figure varies with distance r from the center of the wire according to J = c r , where c is a constant. (a) What is the current through the wire? (b) What is the magnetic field produced by this current for r R ? For r R ?

This figure shows a long, straight, cylindrical wire with a radius R that has current I flowing through it.
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A long, straight, cylindrical conductor contains a cylindrical cavity whose axis is displaced by a from the axis of the conductor, as shown in the accompanying figure. The current density in the conductor is given by J = J 0 k ^ , where J 0 is a constant and k ^ is along the axis of the conductor. Calculate the magnetic field at an arbitrary point P in the cavity by superimposing the field of a solid cylindrical conductor with radius R 1 and current density J onto the field of a solid cylindrical conductor with radius R 2 and current density J . Then use the fact that the appropriate azimuthal unit vectors can be expressed as θ ^ 1 = k ^ × r ^ 1 and θ ^ 2 = k ^ × r ^ 2 to show that everywhere inside the cavity the magnetic field is given by the constant B = 1 2 μ 0 J 0 k × a , where a = r 1 r 2 and r 1 = r 1 r ^ 1 is the position of P relative to the center of the conductor and r 2 = r 2 r ^ 2 is the position of P relative to the center of the cavity.

This figure shows a large circle with a radius R1 that has a circular hole of radius R2 in it at a distance a from the center. Point P is located in a hole at the distance r2 from the center of a hole and at a distance r1 from the center of a large circle.

derivation

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Between the two ends of a horseshoe magnet the field is uniform as shown in the diagram. As you move out to outside edges, the field bends. Show by Ampère’s law that the field must bend and thereby the field weakens due to these bends.

This figure shows a horse shoe magnet with the magnetic lines going from the North end to the South end.
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Show that the magnetic field of a thin wire and that of a current loop are zero if you are infinitely far away.

As the radial distance goes to infinity, the magnetic fields of each of these formulae go to zero.

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An Ampère loop is chosen as shown by dashed lines for a parallel constant magnetic field as shown by solid arrows. Calculate B · d l for each side of the loop then find the entire B · d l . Can you think of an Ampère loop that would make the problem easier? Do those results match these?

This figure shows an Ampere loop that is located in the constant magnetic field. One of the sides of the loop forms an angle theta with the magnetic line.
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A very long, thick cylindrical wire of radius R carries a current density J that varies across its cross-section. The magnitude of the current density at a point a distance r from the center of the wire is given by J = J 0 r R , where J 0 is a constant. Find the magnetic field (a) at a point outside the wire and (b) at a point inside the wire. Write your answer in terms of the net current I through the wire.

a. B = μ 0 I 2 π r ; b. B = μ 0 J 0 r 2 3 R

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A very long, cylindrical wire of radius a has a circular hole of radius b in it at a distance d from the center. The wire carries a uniform current of magnitude I through it. The direction of the current in the figure is out of the paper. Find the magnetic field (a) at a point at the edge of the hole closest to the center of the thick wire, (b) at an arbitrary point inside the hole, and (c) at an arbitrary point outside the wire. ( Hint: Think of the hole as a sum of two wires carrying current in the opposite directions.)

This figure shows a circle with a radius a that has a circular hole of radius b in it at a distance d from the center.
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Magnetic field inside a torus. Consider a torus of rectangular cross-section with inner radius a and outer radius b . N turns of an insulated thin wire are wound evenly on the torus tightly all around the torus and connected to a battery producing a steady current I in the wire. Assume that the current on the top and bottom surfaces in the figure is radial, and the current on the inner and outer radii surfaces is vertical. Find the magnetic field inside the torus as a function of radial distance r from the axis.

B ( r ) = μ 0 N I / 2 π r
This figure shows a torus with the inner radius a and an outer radius b. A thin wire is wound evenly on the torus.

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Two long coaxial copper tubes, each of length L , are connected to a battery of voltage V . The inner tube has inner radius a and outer radius b , and the outer tube has inner radius c and outer radius d . The tubes are then disconnected from the battery and rotated in the same direction at angular speed of ω radians per second about their common axis. Find the magnetic field (a) at a point inside the space enclosed by the inner tube r < a , and (b) at a point between the tubes b < r < c , and (c) at a point outside the tubes r > d . ( Hint: Think of copper tubes as a capacitor and find the charge density based on the voltage applied, Q = V C , C = 2 π ε 0 L ln ( c / b ) .)

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Challenge problems

The accompanying figure shows a flat, infinitely long sheet of width a that carries a current I uniformly distributed across it. Find the magnetic field at the point P, which is in the plane of the sheet and at a distance x from one edge. Test your result for the limit a 0 .

This picture shows a flat, infinitely long sheet of width a that carries a current I uniformly distributed across it. Point P is in the plane of the sheet and at a distance x from one edge.

B = μ 0 I 2 π x .

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A hypothetical current flowing in the z -direction creates the field B = C [ ( x / y 2 ) i ^ + ( 1 / y ) j ^ ] in the rectangular region of the xy -plane shown in the accompanying figure. Use Ampère’s law to find the current through the rectangle.

This figure shows the rectangular region of the xy-plane; z axis is perpendicular to the plane. Points a1 and a2 are located at the x axis. Points b1 and b2 are located at the y axis. There is an equal distance between all points.
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A nonconducting hard rubber circular disk of radius R is painted with a uniform surface charge density σ . It is rotated about its axis with angular speed ω . (a) Find the magnetic field produced at a point on the axis a distance h meters from the center of the disk. (b) Find the numerical value of magnitude of the magnetic field when σ = 1 C/m 2 , R = 20 cm , h = 2 cm , and ω = 400 rad/sec , and compare it with the magnitude of magnetic field of Earth, which is about 1/2 Gauss.

a. B = μ 0 σ ω 2 [ 2 h 2 + R 2 R 2 + h 2 −2 h ] ; b. B = 4.09 × 10 −5 T , 82% of Earth’s magnetic field

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Questions & Answers

Newton's second laws is call with
Dyutee Reply
Really
Arzoodan
what is mean by thermodynamics
Prasad Reply
it is study about temperature and it's equilibrium
thiru
Its the study of heat and its relation with others kind of energy
Antonio
state caulombs law clearly
constand Reply
show mathematically that an electron has the greater speed than the proton when they attract each other
ezra Reply
show mathematically that an electron has the greater speed than the proton when they attract each other
srikanta
@ezra & srikanta; for electrons: a=ke^2/(mr^2) and for protons: a=kp^2/(mr^2)
Sikandar
what is electrostatics
Hero Reply
the study of charge at rest
Gulzar
@Hero; the study of charges at rest is the electrostatics
Sikandar
okay what is electrostatic?
Abd
charge at rest
Nawal
set of character...
Arzoodan
oky
Abd
Gauss law, electric fields, dipoles,...
Antonio
good
Abd
A proton initially at rest falls through a p.d of 25000V. what speed does it gain?
Minister Reply
@Minister; use equation v= sq root(2×eV/m)
Sikandar
what is the reaction of heat on magnet
ORIZINO Reply
Magnetization decreases with increase in temperature. But in case of diamagnetic substance heat has no role on magnetization.
srikanta
what is a physical significant of electric dipole moment .
PRANAB Reply
A dipole moment it's a mechanical electrical effect used in nature
Antonio
what is the uses of carbon brushes in generator
Malik Reply
to minimize heat
constand
at what temperature is the degree Fahrenheit equal to degree Celsius
Grace Reply
Celsius and Faharaneith are different, never equal
Antonio
find their liners express of n=a+b/T² ( plot graph n against T)
Donsmart Reply
Radio Stations often advertis "instant news,,if that meens you can hear the news the instant the radio announcer speaks it is the claim true? what approximate time interval is required for a message to travel from Cairo to Aswan by radio waves (500km) (Assume the waves Casbe detected at this range )
mahmod Reply
what is growth and decay
Pawan Reply
Can someone please predict the trajectory of a point charge in a uniform electric field????
erlinda Reply
what is deference between strong force and coulomb force
zahid Reply
how do you convert temperature in degree Celsius to Fahrenheit
kwame
Celsius x 9/5 +32
Cyclone
Practice Key Terms 6

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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