Magnetism. February 27, 2014 Physics for Scientists & Engineers 2, Chapter 27 1

Transcription

1 Magnetism February 27, 2014 Physics for Scientists & Engineers 2, Chapter 27 1

2 Force on a Current Carrying Wire! The magnitude of the magnetic force on a wire of length L carrying a current i is F = il B B F = ilbsinθ! θ is the angle between the current and the magnetic field! Force is perpendicular to both current and magnetic February 27, 2014 Physics for Scientists & Engineers 2, Chapter 27 2

3 Torque on a Current-Carrying Loop! Electric motors rely on the magnetic force exerted on a current carrying wire! This force is used to create a torque that turns a shaft! A primitive electric motor is depicted below consisting of a square loop of side a carrying current i in a constant magnetic field B February 27, 2014 Physics for Scientists & Engineers 2, Chapter 27 4

4 Torque on a Current-Carrying Loop! The magnitude of the magnetic force on the vertical sections of the loop is F B = iab! The forces on the two vertical sections are in the opposite direction and cause a torque that tends to rotate the loop around the dashed line February 27, 2014 Physics for Scientists & Engineers 2, Chapter 27 5

5 Torque on a Current-Carrying Loop! As the loop rotates, the magnetic forces on the vertical sections do not change! However, the torque changes! We define θ as the angle between the unit normal vector of the loop and the magnetic field! The torque on the loop is ( ) a 2 τ 1 = iab sinθ + iab ( ) a 2! We can see that when θ = 0 there is no torque sinθ = ia 2 Bsinθ = iabsinθ February 27, 2014 Physics for Scientists & Engineers 2, Chapter 27 6

6 Current Creates a Magnetic Field! In Chapter 27 we used a compass to map out the magnetic field of a permanent magnet! If we run a current through a wire and then do the same thing we find the following February 27, 2014 Physics for Scientists & Engineers, Chapter 28 8

7 Reminder: Electric Fields! In Chapter 27 we saw that magnetic field affect moving charges! In this chapter we will show that moving charges create magnetic fields! For an electric field, we showed that the electric field from a differential charge dq is given by d E = 1 4πε 0 dq r 3 r = 1 4πε 0 dq r 2 ˆr! The situation is more complicated for a magnetic field because the current element producing a magnetic field has a direction February 27, 2014 Physics for Scientists & Engineers, Chapter 28 9

8 ! We can write the magnetic field produced by a current element ids as db = µ 0 ids r = µ 0 ids ˆr 4π r 3 4π r 2! This formula is the Biot-Savart Law! μ 0 is the magnetic permeability of free space whose value is µ 0 = 4π 10 7 T m A! The magnitude is given by db = µ 0 i ds sinθ Biot-Savart Law 4π r 2 February 27, 2014 Physics for Scientists & Engineers, Chapter 28 10

9 Magnetic Field from a Long, Straight Wire! Calculate the magnetic field at any point in space as the sum of differential magnetic field using the Biot-Savart Law! First, the magnetic field from a long, straight wire February 27, 2014 Physics for Scientists & Engineers, Chapter 28 11

10 Magnetic Field from a Long, Straight Wire! We will calculate the magnetic field from the right half of the wire and multiply by two to get the magnetic field from the whole wire! The magnitude of the magnetic field from the right side of the wire is given by B = 2 db = 2 0! We can relate the various variables as r = s 2 + r 2 0 µ 0 ids sinθ = µ i 0 4π r 2 2π 0 ds sinθ r 2 sinθ = sin( π θ) = r s 2 + r 2 February 27, 2014 Physics for Scientists & Engineers, Chapter 28 12

11 Magnetic Field from a Long, Straight Wire! Substituting for r and θ gives us B = µ 0 i 2π 0 r ds ( s r ) 3/2! Evaluating this integral gives µ i 1 r s µ i s B = = 0 2π 2 B = µ 0 i 2πr ( 2 2) 1/ 2 ( 2 2) 1/ 2 r s + r π r s + r 0! Math reminders: s (s 2 + x 2 ) = s 1 1/2 s 2 1+ x2 s 2 ( ) February 27, 2014 Physics for Scientists & Engineers, Chapter s = s s x 2 / s 2

12 Magnetic Field from a Long, Straight Wire! The direction is given by another right hand rule February 27, 2014 Physics for Scientists & Engineers, Chapter 28 14

13 Two Parallel Wires! Consider the case in which two parallel wires are carrying currents! The two wires will exert a magnetic force on each other because the magnetic field of one wire will exert a force on the moving charges in the second wire! The magnetic field a distance d from a wire carrying a current i is B = µ 0i 2πd! The direction is given by the right hand rule February 27, 2014 Physics for Scientists & Engineers, Chapter 28 17

14 Two Parallel Wires! Let s start with wire 1 carrying a current i 1 to the right! The magnitude of the magnetic field a distance d from wire 1 is B 1 µ 0i1 = 2π d! Now consider wire 1 carrying a current i 2 in the same direction placed a distance d from wire 1! The magnetic field due to wire 1 will exert a magnetic force on the moving charges in the current flowing in wire 2 February 27, 2014 Physics for Scientists & Engineers, Chapter 28 18

15 ! The magnetic force on a current-carrying wire is F = i L B Two Parallel Wires! The force on a length L of wire 2 is then F = ilbsinθ = i 2 LB 1! Substituting for B 1 gives us the magnitude of the force exerted by wire 1 on a length L of wire 2 F 1 2 = i 2 L µ i 0 1 2πd = µ i i L πd February 27, 2014 Physics for Scientists & Engineers, Chapter 28 19

16 Two Parallel Wires! Right hand rule 1 tells that the force on wire 2 points toward wire 1! We can redo the same calculation to get the force exerted by wire 1 on wire 2! We find that the force exerted by wire 1 on wire 2 has the same magnitude and is in the opposite direction! Newton s Third Law February 27, 2014 Physics for Scientists & Engineers, Chapter 28 20

17 Magnetic Field due to a Wire Loop! Using more advanced techniques and with the aid of a computer we can determine magnetic field from a wire loop at other points in space February 27, 2014 Physics for Scientists & Engineers, Chapter 28 22

18 Ampere s Law! We have calculated the magnetic field from an arbitrary distribution of current elements using d B = µ 0 4π! We may be faced with a difficult integral! In cases where the distribution of current elements has cylindrical or spherical symmetry, we can apply Ampere s Law to calculate the magnetic field from a distribution of current elements with much less effort than using a direct integration! Ampere s Law is B d s id s r r 3 = µ 0 i enc February 27, 2014 Physics for Scientists & Engineers, Chapter 28 23

19 Magnetic Fields of Solenoids! Current flowing through a single loop of wire produces a magnetic field that is not very uniform! Applications often require a uniform field! A first step toward a more uniform magnetic field is the Helmholtz coil! This coil consists of 2 sets of coaxial wire loops February 27, 2014 Physics for Scientists & Engineers, Chapter 28 24

20 Magnetic Fields of Solenoids! Let s look at the magnetic field created by 4 coaxial coils! Let s go from 4 coils to many coils February 27, 2014 Physics for Scientists & Engineers, Chapter 28 25

21 Magnetic Fields of Solenoids! Let s look at the magnetic field created by 600 coaxial coils! 600 coils produce an excellent field inside the coils February 27, 2014 Physics for Scientists & Engineers, Chapter 28 26