Phys102 Lecture 16/17 Ampere's Law

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Presentation transcript:

Phys102 Lecture 16/17 Ampere's Law Key Points Ampère’s Law Magnetic Field Due to a Straight Wire B Magnetic Field of a Solenoid References 20-5,6,7.

Ampère’s Law Ampère’s law relates the magnetic field around a closed loop to the total current flowing through the loop: where Figure 28-8. Arbitrary path enclosing a current, for Ampère’s law. The path is broken down into segments of equal length Δl. μ0 = 4  10-7 T∙m/A is the magnetic permeability of free space.

Using Ampère’s law to find the field around a long straight wire: Use a circular path with the wire at the center; then B is tangent to Δl at every point. Then Figure 28-9. Circular path of radius r. Similar to Gauss’s law, we need symmetry.

i-clicker 1. Magnetic Field A) + y B) – y C) + x D) + z (out of page) E) – z (into page) A proton beam enters a magnetic field region as shown below. What is the direction of the magnetic field B? x y

i-clicker 2 Magnetic Force on a Loop If there is a current in the loop in the direction shown, the loop will: A) move up B) move down C) rotate clockwise D) rotate counterclockwise E) both rotate and move N S B field out of North B field into South

Force between Two Parallel Wires The magnetic field produced at the position of wire 2 due to the current in wire 1 is The force this field exerts on a length l2 of wire 2 is Figure 28-5. (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.

Force between Two Parallel Wires Parallel currents attract; antiparallel currents repel. Figure 28-6. (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.

Example. Force between two current-carrying wires. The two wires of a 2.0-m-long appliance cord are 3.0 mm apart and carry a current of 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.

Definitions of the Ampere and the Coulomb 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 2 x 10-7 N per meter of length of each wire. The coulomb is then defined as exactly one ampere-second.

Magnetic Field of a Solenoid A solenoid is a coil of wire containing many loops. To find the field inside, we use Ampère’s law along the path indicated in the figure. Figure 28-16: Cross-sectional view into a solenoid. The magnetic field inside is straight except at the ends. Red dashed lines indicate the path chosen for use in Ampère’s law. (n is the number of loops per unit length)

Example: Field inside a solenoid. A thin 10-cm-long solenoid used for fast electromechanical switching has a total of 400 turns of wire and carries a current of 2.0 A. Calculate the field inside near the center. Solution: Substitution gives B = 1.0 x 10-2 T.

Magnetic Fields in Magnetic Materials* If a ferromagnetic material such as iron is placed in the core of a solenoid, the magnetic field is enhanced by the field created by the ferromagnet itself. This is usually much greater than the field created by the current alone. The magnetic field can be written as B = μnI where μ is the magnetic permeability, ferromagnets have μ >> μ0 (e.g. μ ≈ 2000 μ0 for electrical steel), while all other materials have μ ≈ μ0.