Download presentation

Presentation is loading. Please wait.

Published byMichelle Whitaker Modified over 3 years ago

1
Magnetism Part II Field and Flux

2
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

3
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.

4
Biot-Savart Law dB = (μ o /4π)[(Ids x r)/r 2 ] μ o = 4π x 10 -7 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

5
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

6
I a P x r θ

7
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)

8
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.

9
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

10
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

11
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

12
HW Contd Ch. 30 prob. # 8,16 and 20

Similar presentations

Presentation is loading. Please wait....

OK

Calculating the Magnetic Field Due to a Current

Calculating the Magnetic Field Due to a Current

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on meeting skills training Ppt on indian english literature Ppt on chapter human eye and the colourful world Ppt on p&g products msds sheets Ppt on right to information act Ppt on boilers operations with integers Download ppt on web browser Ppt on kpo industry in india Ppt on coalition government in uk Free download ppt on albert einstein