Ampere question

# Ampere question - Biot Savat and Ampere Last time we said...

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Biot Savat and Ampere Last time we said that the magnetic field produced at a distance r by a charge q moving with a velocity v was B =(μ o q)/ (4 π r 2 ) ( v x r-hat ) Now the magnetic field produced by a current flowing in a wire is of much greater practical interest than that produced by a single moving charge. A current in a wire is a distribution of moving point charges, each of which produces its own individual magnetic field. We can add all these little field up to get the net field. dl r-hat I ø The figure shows a thin wire of arbitraty shape carrying a steady current I, which, as usual, we pretend is positive. Consider a small segment dl of the wire. The moving charge dq within dl produces a magnetic field dB dB =(μ o dq)/ (4 π r 2 ) ( v x r-hat ) and this is going to have a magnitude dB =(μ o dq)/ (4 π r 2 ) v sin ø where ø is the angle between v and r now dq = I dt

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and v = dl/dt so dt = dl/v dB =(μ o Idt)/ (4 π r 2 ) v sin ø dB =(μ o I dl/v)/ (4 π r 2 ) v sin ø
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## This note was uploaded on 01/09/2010 for the course PHYS 221 taught by Professor G.r.grist during the Spring '09 term at Skyline College.

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Ampere question - Biot Savat and Ampere Last time we said...

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