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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.
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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.
Example: Using Ampère’s Lawto 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.
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.
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.
Example A “nice use” for Ampère’s Law. Use Ampère’s Law to show that in any region of space where there are no currents the magnetic field cannot be both unidirectional & nonuniform as shown in the figure.
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.
Magnetic Field of a Solenoid & 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 = 0outside the solenoid, & the path integral is zero along the vertical lines, so the field is (n = number of loops per unit length):
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.
Biot-Savart Law The Biot-Savart Law calculates the vector magnetic field dB due to an infinitesimal length of current. The total vector field B can be found by integrating over the total length of all currents:
Example: B Due to a Current I in Straight Wire. For the magnetic field near a long straight wire carrying a current I, show that the Biot-Savart Law gives the same magnetic field as Ampère’s Law: B = (μ0I)/(2πr)
Example: Current Loop CalculateB for points on the axis of a circular loop of wire of radius R carrying a current I.
Example:B Due to a Wire Segment. One quarter of a circular loop of wire carries a current I. The current I enters & leaves on straight segments of wire, as shown; the straight wires are along the radial direction from the center C of the circular portion. Calculate the magnetic field at point C.
