Download presentation

Presentation is loading. Please wait.

1
**Magnetic Fields Due to Currents**

Magnetism Part II Magnetic Fields Due to Currents

2
Ampere’s Law The line integral of the magnetic field B around any closed path is equal to μo times the net current across the area bound by the path. ∫ B dl = μoI (integral around a closed loop)

3
Using Ampere’s Law Select the integration path. If you want to determine B at a point, then the path must pass through the point. The integration path doesn’t have to be any actual physical boundary. Usually it is a purely geometric curve. The integration path has to have enough symmetry to make evaluation of the integral possible.

4
Continued The B is tangent to all or some portion of the integration path and has the same magnitude of B at every point, then its integral equals Bl for that portion of the path. If B is perpendicular to all or some portion of the path, that portion makes no contribution to the integral.

5
Continued The sign of a current enclosed by the integration path is given by the right hand rule. Curl your fingers of your right hand so that they follow the integration path in the direction you carry out the integration. Your right thumb then points in the direction of positive current. If B is as described in step 4 and I enclosed is positive, then the direction of B is the same as the direction of the integration path: if instead I enclosed is negative, B is in the direction opposite that of integration.

6
Continued In the integral, B is always the total magnetic field at each point on the path. This field can be caused partly by currents enclosed by the path and partly by currents outside. If the path encloses no net current, the field points on the path need not be zero, but the integral is always zero.

7
Long Straight Wire A long straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire. Calculate the magnetic field a distance r from the center of the wire in the region r ≥ R and r < R

8
Solution r ≥ R ∫ B dl = B ∫dl = B (2πr) = μoI B = μoI/(2πr)

9
**Solution r < R I’<I I’/I = π r2/πR2 I’ = (r2/R2) I**

∫ B dl = B(2 πr) = μoI’ B(2 πr) = μo(r2/R2)I B = μorI/(2πR2)

10
Toroids

11
Toroid A toroid (wire segment around a non-conducting ring) has N closely spaced turns of wire, calculate the magnetic field in the region occupied by the torus a distance r from its center.

12
Solution ∫B dl = μo I B ∫dl = μo NI B (2πr) = μo NI B = (μo NI)/(2πr)

13
Problem A tightly wound solenoid has a length of 30 cm, a diameter of 2.0 cm, and contains 10,000 turns. If it carries a current of 5A, what is the magnitude of the magnetic field inside the solenoid?

14
Solution B = μo nI B = μo (N/L)I B = 0.2T

Similar presentations

OK

The Biot-Savart Law. Biot and Savart recognized that a conductor carrying a steady current produces a force on a magnet. Biot and Savart produced an equation.

The Biot-Savart Law. Biot and Savart recognized that a conductor carrying a steady current produces a force on a magnet. Biot and Savart produced an equation.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on limits and derivatives for class 11 Ppt on acid base and salts Ppt on job evaluation point Ppt on different types of dance forms in africa Ppt on job rotation advantages Working of raster scan display ppt on tv Dsp ppt on dft communications Ppt on business environment nature concept and significance of the study Ppt on print media and electronic media Ppt on history of world wide web