1. AP Physics ST Solenoid Magnetic Flux Gauss’s Law in Magnetism

2. Solenoid Solenoid Long wire wound to form a helix.
A current-carrying wire that produces a decently uniform magnetic field along axis.
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3. Solenoid Loosely wound coil Notice B relatively uniform inside coil.
B weak outside coil.
Notice
Significant magnetic field generated between each wire; sloppy and weak field outside coil.
Non-uniformity of field inside coil.

4. Solenoid Tightly wound coil Notice B very uniform inside coil.
When tightly wound each turn can be approximated as a circular loop.
Increase the length of solenoid… more uniform B becomes.
B negligible outside coil.
Notice
Magnetic field between each wire more effectively cancels out.
Significantly greater uniformity of field inside coil.

5. Solenoid Between wires: Internal region of coil:
Magnetic field between every two wires cancel each other.
Internal region of coil:
All magnetic field contributions directed UP.
External region of coil:
All magnetic field contributions directed DOWN. B B I B B B

6. Solenoid Consider an ideal solenoid Turns are closely spacedB I W
Consider an ideal solenoid
Turns are closely spaced
Length long compared to its radius of curvature
Magnetic field is uniform inside the coil
Current out of the page on lefts side of coil, into the page on the right side of coil 2 1 3 L 4

7. Solenoid Apply Ampere’s Law to each region along the path from 1 to 4.B I W
Apply Ampere’s Law to each region along the path from 1 to 4.
Note that ds, path element is represented by L in this situation. 2 1 3 L 4

8. Solenoid Leg 3… Leg 2,4… Leg 1 …
No magnetic field passing along this leg; outside solenoid.
Leg 2,4…
B is perpendicular to ds therefore dot-product equals zero.
Leg 1 …
Magnetic field parallel to ds therefore this leg WILL contribute.

9. Solenoid
Summing “B dot ds” contributions along length L in presence of B…
Total current passing along length L…
N = # loops

10. Solenoid Apply Ampere’s Law… Substitute contributing elements…
Magnetic field of a Solenoid…
n = # loops per length (of solenoid)

11. Magnetic Flux Recall the concept of electric flux: Magnetic Flux:
Proportional to the number of electric field lines penetrating an area element.
Magnetic Flux:
Proportional to the number of magnetic field lines penetrating an area element…

12. Magnetic Flux Minimum flux: Maximum flux: B perpendicular to dA…
B parallel to dA…
surface B B dA
surface

13. Magnetic Flux Flux at any relative angle between dA and B…
Consider the component of dA in the direction of B. dA θ B
surface

14. Gauss’s Law in Magnetism
Recall electric flux:
The number of electric field lines leaving a gaussian surface is proportional to the charge enclosed by the gaussian surface.
Note the electric field lines are only leaving the gaussian surface!!
electricfield.info

15. Gauss’s Law in Magnetism
Magnetic flux:
The number of magnetic field lines leaving a gaussian surface is proportional to the magnetic field enclosed by the gaussian surface.
But the magnetic field lines both enter then leave the surface at all points!! WHAT DOES THIS MEAN????

16. Gauss’s Law in Magnetism
If magnetic field lines both enter and leave every surface then
The number of lines entering equal the number of lines leaving.
Magnetic field line always leave a N pole and end on a S pole forming continuous loops.
The magnetic flux is ALWAYS ZERO!!!

17. Gauss’s Law in Magnetism
Gauss’s Law of Magnetism…

18. Lesson Summary

19. Lesson Summary

20. Example #1 Magnetic Fluxdr
A long straight wire carries a current I and produces a magnetic field B. Determine the magnetic flux through a region of width a and length b a distance c from the wire. I r b c a

21. Example #2 Problem 30-32
A single-turn square loop of wire with an edge length of 2 cm carries a clockwise current of A. the loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns/cm and carries a clockwise current of 15 A.
Find the force on each side of the loop.
Find the torque acting on the loop.
Isqr = 0.20 A
0.02 m
0.02 m
Isol = 15 A