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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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**Express x in terms of the angle and find dx.**

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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AP Physics C Montwood High School R. Casao

AP Physics C Montwood High School R. Casao

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