L11 - Chapter 29 continued .... Ampere's Law Andre Marie...

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Unformatted text preview: Chapter 29 continued .... Ampere's Law Andre Marie Ampere (1775-1836) Ampere's Law r v B ds = oiencl Enclosing a current carrying wire with a loop and taking the dot product of B and ds along that loop determines the amount of current enclosed by the loop. For geometries with a high degree of symmetry one can determine the magnetic field from Ampere's LAW Ampere's Law How do you use this equation to determine B? r v B ds = oiencl Examples of Current sources with a high degree of symmetry Long Straight Wire Toroid Solenoid 1) Long Straight Wire with a constant current For r>R, radius of wire the magnetic field must have cylindrical symmetry r v B ds = oienc B ds = oienc B 2r = oi r oi ^ t B= 2r ^ unit vecto r tangent to t Amperian loop ^ to be compared to n Example Long Straight Wire r v B ds = oienc 2 B ds = o Jr o i B ds = r R 2 2 i 2 B 2r = o 2 r R For r < R the magnetic field must have r o i r ^ cylindrical symmetry and the current B = t enclosed is only part of I 2 R 2 Example Long Straight Wire Result of considering paths 1 and 2 Sample Problem What is the magnetic field at a=0cm, b=2.0cm a) r=0 cm b) r=1.3 cm J=cr2 Solenoid - A long cylinder with wire closely wrapped around it. The length is much larger than the diameter. Ideal Solenoid r v B ds = oienc r r r r B ds + B ds a r r r r + B ds + B ds = o ienc c d b Bh + 0 + 0 + 0 = oienc Bh = o N turns i where N = nh B = o ni Constant field in x-direction r B = o ni ^ x Magnetic Field inside a donut (toroid) mmm ...... r v B ds = oienc B ds = o Ni B 2r = o Ni r o Ni ^ B= t 2r There is "circular symmetry" here, the field will be constant along the circular path concentric with the center axis of Example Problem 5 What current must flow through the wires of a toroid to generate a 1.0 T magnetic field at its center (10 cm radius) if it has 500 turns of wire? etc ..... o Ni B= 2r ...
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