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**Chapter 30: Sources of the Magnetic Field**

Chapter 28 opener. A long coil of wire with many closely spaced loops is called a solenoid. When a long solenoid carries an electric current, a nearly uniform magnetic field is produced within the loops as suggested by the alignment of the iron filings in this photo. The magnitude of the field inside a solenoid is readily found using Ampère’s law, one of the great general laws of electromagnetism, relating magnetic fields and electric currents. We examine these connections in detail in this Chapter, as well as other means for producing magnetic fields.

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**Chapter Outline Definitions of the Ampere & the Coulomb Ampère’s Law**

Magnetic Field Due to a Straight Wire Force between Two Parallel Wires Definitions of the Ampere & the Coulomb Ampère’s Law Magnetic Field of a Solenoid and a Toroid Biot-Savart Law Magnetic Materials – Ferromagnetism Electromagnets and Solenoids – Applications Magnetic Fields in Magnetic Materials; Hysteresis Paramagnetism and Diamagnetism

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**Magnetic Field Due to a Long, Straight Current Carrying Wire: Direction**

Ch. 29: The magnetic field lines for a long, straight, current carrying wire are circles concentric with the wire. The field lines are in planes perpendicular to the wire. The magnitude of the field is constant on a circle of radius a. Use the right-hand rule to determine the direction of the field, as shown.

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**Magnetic Field Due to a Current Carrying Wire**

A compass can be used to detect the magnetic field. When there is no current in the wire, there is no field due to the current. In this case, the compass needles all point toward the Earth’s north pole. Due to the Earth’s magnetic field

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**Magnetic Field Due to a Current Carrying Wire**

When the wire carries a current, the compass needles deflect in a direction tangent to the circle. This shows the direction of the magnetic field produced by the wire. If the current is reversed, the direction of the needles also reverses.

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**The circular magnetic field around the wire is shown by the iron filings.**

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**Magnetic Field Due to a Straight Wire**

Experimental Results show that the Magnetic field B due to a straight, current carying wire is proportional to the Current I & inversely proportional to the distance r from the wire: μ0 is a constant, called the permeability of free space. It’s value is μ0 = 4π 10-7 T·m/A. It plays a similar role for magnetic fields that ε0 plays for electric fields! Figure Same as Fig. 27–8b. Magnetic field lines around a long straight wire carrying an electric current I.

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**Calculation of B Near a Wire**

An electric wire in the wall of a building carries a dc current I = 25 A vertically upward. Calculate the magnetic field B due to this current at a point P = 10 cm in the radial direction from the wire. Solution: B = μ0I/2πr = 5.0 x 10-5 T

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**Calculation of B Near a Wire**

An electric wire in the wall of a building carries a dc current I = 25 A vertically upward. Calculate the magnetic field B due to this current at a point P = 10 cm in the radial direction from the wire. Solution Use This gives B = 5.0 10-5 T Solution: B = μ0I/2πr = 5.0 x 10-5 T

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**Magnetic field midway between two currents.**

Example Magnetic field midway between two currents. Two parallel straight wires a distance 10.0 cm apart carry currents in opposite directions. Current I1 = 5.0 A is out of the page, and I2 = 7.0 A is into the page. Calculate the magnitude & direction of the magnetic field halfway between the 2 wires. Figure Example 28–2. Wire 1 carrying current I1 out towards us, and wire 2 carrying current I2 into the page, produce magnetic fields whose lines are circles around their respective wires. Solution: As the figure shows, the two fields are in the same direction midway between the wires. Therefore, the total field is the sum of the two, and points upward: B1 = 2.0 x 10-5 T; B2 = 2.8 x 10-5 T; so B = 4.8 x 10-5 T.

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**Magnetic field midway between two currents.**

Example Magnetic field midway between two currents. Two parallel straight wires a distance 10.0 cm apart carry currents in opposite directions. Current I1 = 5.0 A is out of the page, and I2 = 7.0 A is into the page. Calculate the magnitude & direction of the magnetic field halfway between the 2 wires. Solution: Use for each wire. Then add them as vectors B = B1 + B2 B = 4.8 10-5 T Figure Example 28–2. Wire 1 carrying current I1 out towards us, and wire 2 carrying current I2 into the page, produce magnetic fields whose lines are circles around their respective wires. Solution: As the figure shows, the two fields are in the same direction midway between the wires. Therefore, the total field is the sum of the two, and points upward: B1 = 2.0 x 10-5 T; B2 = 2.8 x 10-5 T; so B = 4.8 x 10-5 T.

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**Magnetic Field Due to 4 Wires**

Conceptual Example Magnetic Field Due to 4 Wires The figure shows 4 long parallel wires which carry equal currents into or out of the page. In which configuration, (a) or (b), is the magnetic field greater at the center of the square? a b Solution: Solution: S: S: Solution: The fields cancel in (b) but not in (a); therefore the field is greater in (a).

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**Magnetic Field Due to 4 Wires**

Conceptual Example Magnetic Field Due to 4 Wires The figure shows 4 long parallel wires which carry equal currents into or out of the page. In which configuration, (a) or (b), is the magnetic field greater at the center of the square? a b Solution: The fields cancel in (b) but not in (a); therefore the field is greater in (a).

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**Magnetic Force Between Two Parallel Wires**

Recall from Ch. 29 that the force F on a wire carrying current I in a magnetic field B is: We’ve just seen that the magnetic field B produced at the position of wire 2 due to the current in wire 1 is: Figure (a) Two parallel conductors carrying currents I1 and I2. (b) Magnetic field B1 produced by I1 (Field produced by I2 is not shown.) B1 points into page at position of I2. The force F this field exerts on a length ℓ2 of wire 2 is The magnetic force between 2 currents is analogous to the Coulomb electric force between 2 charges!!

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**Parallel Currents ATTRACT Anti-Parallel Currents REPEL.**

Using the cross product form of the force: & applying the right hand rule shows that Parallel Currents ATTRACT and Anti-Parallel Currents REPEL. Figure (a) Parallel currents in the same direction exert an attractive force on each other. (b) Antiparallel currents (in opposite directions) exert a repulsive force on each other.

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**Example Force Between Two Current-carrying Wires.**

The two wires of a 2.0-m long appliance cord are d = 3.0 mm apart & carry a current I = 8.0 A dc. Calculate the force one wire exerts on the other. Solution: F = 8.5 x 10-3 N, and is repulsive. Appliances powered by 12V dc current are marketed to truckers, campers, people running on solar power (especially off-grid), among others.

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**Example Force Between Two Current-carrying Wires.**

The two wires of a 2.0-m long appliance cord are d = 3.0 mm apart & carry a current I = 8.0 A dc. Calculate the force one wire exerts on the other. Use: Resulting in F = 8.5 10-3 N Solution: F = 8.5 x 10-3 N, and is repulsive. Appliances powered by 12V dc current are marketed to truckers, campers, people running on solar power (especially off-grid), among others.

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**Example Suspending a wire with a current.**

A horizontal wire carries a current I1 = 80 A dc. Calculate the current I2 that a second parallel wire d = 20 cm below it must carry so that it doesn’t fall due to gravity. The lower wire has a mass per meter of (m/ℓ) = 0.12 g/m. Solution: The magnetic force due to the current in the wire must be equal and opposite to the gravitational force on the second wire. We need a definite length of wire, as the forces on an infinitely long wire will be infinite; choose 1 meter for simplicity. Substitution gives I2 = 15 A.

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**Definitions of the Ampere & the Coulomb**

The SI unit of current, the Ampere is officially defined in terms of the force between two current-carrying wires: One Ampere IS DEFINED as that current flowing in each of two long parallel wires 1 m apart, which results in a force of exactly F =2 10-7 N per meter of length of each wire. One Coulomb is IS DEFINED as exactly 1 ampere-second.

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