Presentation on theme: "Magnetic Field Along the Axis of a Solenoid"— Presentation transcript:
1 Magnetic Field Along the Axis of a Solenoid AP Physics CMontwood High SchoolR. 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.
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.