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The Magnetic Field of a Solenoid AP Physics C Montwood High School R. Casao.

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Presentation on theme: "The Magnetic Field of a Solenoid AP Physics C Montwood High School R. Casao."— Presentation transcript:

1 The Magnetic Field of a Solenoid AP Physics C Montwood High School R. Casao

2 A solenoid is a long wire wound as a helix to produce a reasonably uniform magnetic field B in the interior of the solenoid coils when the solenoid carries a steady current I. When the coils (turns) are closely spaced, each turn can be approximated as a circular loop and the net magnetic field B is the vector sum of the fields resulting from all of the turns.

3 Inside the solenoid, the field lines are nearly parallel, uniformly distributed, and close together, indicating that the magnetic field is uniform. The magnetic field lines between the turns tends to cancel each other. The magnetic field outside the solenoid is nonuniform and weak.

4 The field at exterior points, such as P, are weak because current elements on the upper portions tends to cancel the current elements on the lower portions.

5 If the turns are closely spaced and the solenoid is of a finite length, the field lines resemble the magnetic field lines of a bar magnet, with field lines diverging from one end and converging at the opposite end.

6 One end of the solenoid behaves like the north pole of a bar magnet (diverging end) and the other end behaves like a south pole (converging end). As the length of the solenoid increases, the magnetic field within it becomes more and more uniform. An ideal solenoid is one in which the turns are closely spaced and the length is long compared to the radius. –The magnetic field outside the solenoid is weak compared to the magnetic field inside the solenoid –The magnetic field is uniform over a large volume.

7 Apply Ampere’s law to determine the magnetic field inside an ideal solenoid. A longitudinal cross section of part of the ideal solenoid carries a current I. The magnetic field B inside the solenoid is uniform and parallel to the axis of the solenoid and the magnetic field B outside the solenoid is zero.

8 s Consider an Amperian rectangle of length l and width w. Apply Ampere’s law by evaluating the integral of Bds over each of the four sides of the rectangle. The contribution from side 3 is zero because B = 0 T outside the solenoid.

9 The contributions from sides 2 and 4 are both 0 since B is perpendicular to ds along these paths and Bds = B·ds·cos  = B·ds·cos 9  Along side 1, B is parallel to the length l, therefore, all elements of length ds are parallel to B and Bds = B·ds·cos  = B·ds·cos  B·ds 

10 Mathematically in integral form: The right side of Ampere’s law involves the total current that passes through the area bound by the rectangular path of integration. The total current through the rectangular path (I enclosed ) equals the current through each turn multiplied by the number of turns.

11 Ampere’s law: The quantity is the number of turns per unit length: Magnetic field B inside a solenoid:

12 The magnetic field equation works best for points near the center of a long solenoid. –The magnetic field near each end of the solenoid is smaller than the value given by the equation. –At the end of a long solenoid, the magnitude of the magnetic field B is about one half that of the field near the center. The direction of the magnetic field in a solenoid can be determined using the right hand rule: –Curl the fingers of the right hand in the direction of the current. –The thumb points in the direction of the magnetic field.

13 Rectangular Toroid For problems that reference a rectangular toroid: –The toroid shape remains circular. –The interior area is a rectangle of inner radius a and outer radius b and height h.

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