# Magnetic Field Along the Axis of a Solenoid

## Presentation on theme: "Magnetic Field Along the Axis of a Solenoid"— Presentation transcript:

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

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.

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.

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:

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

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.

Substitute:

Integrate from 1 to 2:

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.

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.

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