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Published byClaire Reynolds Modified over 3 years ago

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Magnetic Field Along the Axis of a Solenoid AP Physics C Montwood High School R. Casao

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Consider a solenoid of length L, radius R, turns N, carrying current I. We will determine an equation for the magnetic field B at an axial point P inside the solenoid.

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Consider the solenoid as a distribution of current loops. The magnetic field for any one loop is: The net magnetic field in the solenoid is the sum of the magnetic fields of all the loops.

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Divide the length of the solenoid into small elements of length dx. –The number of turns in a length dx is: –The amount of current in an element of length dx is: –The total current in a length dx is:

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The magnetic field contribution dB at point P due to each element dx carrying current di is:

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For each element of length dx along the length of the solenoid, the distance x and the angle change. –The value of R remains constant. Express x in terms of the angle and find dx.

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Substitute:

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Integrate from 1 to 2 :

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If point P is at the midpoint of the solenoid and if the solenoid is long in comparison to the radius R, then 1 = -90° and 2 = 90°. The result is the equation for the magnetic field at the center of a solenoid.

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If point P is a point at the end of a long solenoid towards the bottom, then 1 = 0° and 2 = 90°. The answer shows that the magnetic field at the end of a solenoid approaches ½ the value at the center of the solenoid.

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Graph of magnetic field B at axial points vs. distance x for a solenoid.

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