Download presentation
1
Magnetic Field Along the Axis of a Solenoid
AP Physics C Montwood High School R. Casao
2
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.
3
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.
4
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:
5
The magnetic field contribution dB at point P due to each element dx carrying current di is:
6
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.
7
Substitute:
10
Integrate from 1 to 2:
11
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.
12
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.
13
Graph of magnetic field B at axial points vs. distance x for a solenoid.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.