Presentation on theme: "Magnetism Part II Field and Flux. Origins of Magnetic Fields Using Biot-Savart Law to calculate the magnetic field produced at some point in space by."— Presentation transcript:
Magnetism Part II Field and Flux
Origins of Magnetic Fields Using Biot-Savart Law to calculate the magnetic field produced at some point in space by small current elements. Using Amperes Law to calculate the magnetic field of a highly symmetric configuration carrying a steady current
Biot-Savart Law The vector dB is perpendicular both to ds and to the unit vector r directed from ds to P. The magnitude of dB is inversely proportional to r 2, where r is the distance from ds to P The magnitude of dB is proportional to the current and to the magnitude ds of the length element ds. The magnitude of dB is proportional to sin θ, where θ is the angle between the vectors ds and r.
Biot-Savart Law dB = (μ o /4π)[(Ids x r)/r 2 ] μ o = 4π x Tm/A B = (μ o I/4π) (ds x r)/r 2 The integral is taken over the entire current distribution The magnetic field determined in these calculations is the field created by a current- carrying conductor This can be used for moving charges in space
Problem Consider a thin, straight wire carrying a constant current I and placed along the x axis as shown. Determine the magnitude and direction of the magnetic field at point P due to this current. I P a
I a P x r θ
Solution ds x r = k(ds x r) = k (dx sin θ) k being the unit vector pointing out of the page dB = dB k = (μ o I/4π) [(dx sin θ)/r 2 ]k sin θ = a/r so r = a/ sin θ = a csc θ x = -a cot θ dx = a csc 2 θ dB = (μ o I/4π) (a csc 2 θ sin θ d θ)/(a 2 csc 2 θ) B =(μ o I/4πa) sin θ d θ = (μ o I/4πa) cos θ from θ = 0 to θ= π B = (μ o I/2πa)
Problem Calculate the magnetic field at point O for the current-carrying wire segment shown. The wire consists of two straight portions and a circular arc of radius R, which subtends and angle θ. The arrow heads on the wire indicate the direction of the current.
Solution The magnetic field at O due to the current in the straight segments is zero because ds is parallel to r along these paths In the semicircle ds is perpendicular to r so ds x r is ds O A A C C
Solution Continued dB = (μ o I/4π) ( ds/R 2 ) B = (μ o I/4π)/R 2 ds B = [(μ o I/4π)s]/R 2 s = rθ B = (μ o I/4πR) θ
HW Problem Consider a circular loop of radius R located on the yz plane and carrying steady current I. Calculate the magnetic field at an axial point P a distance x from the center of the loop. y z P I