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Ampère’s Law

Ampère’s Law. Amp è re’s Law relates the magnetic field B around a closed loop to the total current I encl flowing through the loop:. The integral is taken around the outside edge of the closed loop. Example : Use Amp è re’s Law to find the field around a

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Ampère’s Law

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  1. Ampère’s Law

  2. Ampère’s Law relates • the magnetic field B • around a closed loop to • the total current Iencl • flowing through the loop: • The integral is taken around the • outside edge of the closed loop.

  3. Example: Use 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 dl at every point. The • integral then gives: so B = (μ0I)/(2πr),as before.

  4. Example: Field Inside & Outside a Current Carrying Wire • A long, straight cylindrical wire • conductor of radius R carries a current • I of uniform current density in the • conductor.Calculate the magnetic • field due to this current at • (a) points outside the conductor (r > R) (b) points inside the conductor (r < R). • Assume that r, the radial distance from the • axis, is much less than the length of the wire. • (c) If R = 2.0 mm & I = 60 A, • CalculateB at r = 1.0 mm, • r = 2.0 mm, & r = 3.0 mm.

  5. Field Due to a Long Straight WireFrom Ampere’s Law • Outside of the wire, r > R: So B = (μ0I)/(2πr), as before.

  6. Here, we need Iencl, the current inside the Amperian loop. The current is uniform, so Iencl= (r2/R2)I so B(2r) = μ0(r2/R2)I which gives: B = (μ0Ir)/(2R2) • Inside of the wire, r < R:

  7. Field Due to a Long Straight Wire: Summary • The field isproportional to r inside the wire. B = (μ0Ir)/(2R2) • The fieldvaries as1/r outside the wire. B = (μ0I)/(2πr)

  8. Conceptual Example: Coaxial Cable. • A Coaxial Cable is a single wire surrounded by a • cylindrical metallic braid. The 2 conductors are separated • by an insulator. The central wire carries current to the other • end of the cable, & the outer braid carries the return current & • is usually considered ground. Calculatethe magnetic field • (a) in the space between the conductors • (b) outside the cable.

  9. Solving Problems Using Ampère’s Law • Ampère’s Lawis most useful for solving • problems when there is considerable Symmetry. • Identify the Symmetry. • Choose an Integration Path that reflects • the Symmetry(typically, the best path is along • lines where the field is constant & perpendicular • to the field where it is changing). • Use the Symmetryto determine the direction of • the field. • Calculate the Enclosed Current.

  10. Magnetic Field of a Solenoid & of a Toroid Solenoid A coil of wire with many loops. To find the field inside, use Ampère’s Law along the closed path in the figure. B = 0 outside the solenoid, & the path integral is zero along the vertical lines, so the field is (n = number of loops per unit length):

  11. Example: Field Inside a Solenoid A thin ℓ = 10 cm long solenoid has a total of N = 400 turns (n = N/ℓ) of wire & carries a current I = 2.0 A. Calculate the magnetic field inside near the center. A Toroid is similar to a solenoid, but it is bent into the shape of a circle as shown. Example: Toroid Use Ampère’s Law to calculate the magnetic field (a) inside & (b) outside a toroid.

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