Magnetic Fields due to Currents

Transcription

1 s s Water, fire, air and dirt, [freaking] magnets, how do they work? - Insane Clown Posse David J. Starling Penn State Hazleton PHYS 212

2 Moving charges are affected by magnetic fields: F B = q v B But there is a symmetry to this force/field relationship Moving charges also create magnetic fields An empirical result: d B = µ 0 i d s ˆr 4π r 2 (Biot-Savart Law) s

3 d B = µ 0 i d s ˆr 4π r 2 µ 0 : permeability of free space or magnetic constant µ 0 = m kg/s 2 A 2 s r is distance between current and point of interest ˆr points from current to point of interest d s points along current direction

4 A long wire: What is the field from a long wire with current i? s

5 s Let s do the math: Start with Biot-Savart law and fill in the gaps: db = µ 0 i ds sin(θ) 4π r 2 r = R 2 + s 2 sin(θ) = sin(π θ) = O/H = R/r = R/ R 2 + s 2 db = µ 0 ir ds 4π (s 2 + R 2 ) 3/2

6 s This field is from a small part of the wire: db = µ 0 ir ds 4π (s 2 + R 2 ) 3/2 But this is the total magnetic field: B = µ 0iR 4π = µ 0i 4πR s= = µ 0i [1 + 1] 4πR = µ 0i 2πR ds s= (s 2 + R 2 ) [ ] 3/2 s (s 2 + R 2 ) 1/2 The direction is given by the right-hand-rule.

7 What about an arc of wire? s The total magnetic field: B = µ 0 i ds sin(θ) 4π r 2 ds = R dφ sin(θ) = sin(π/2) = 1 R = constant i = constant

8 What about an arc of wire? s The total magnetic field: B = µ φ 0i 4π = µ 0i 4πR = µ 0iφ 4πR 0 φ 0 R dφ sin(π/2) dφ R 2

9 Lecture Question 12.1: How does the result of the previous calculation change if we include current in sections 1 and 2 in the figure below? s (a) The magnetic field is larger. (b) The magnetic field is smaller. (c) The magnetic field is the same. (d) The magnetic field changes direction.

10 How do two currents affect each other? s The first current creates a magnetic field: B a = µ 0i a 2πd The second current feels a force from this magnetic field: F ba = i b LB a = µ 0Li a i b 2πd

11 s How do two currents affect each other? F ba = i b LB a = µ 0Li a i b 2πd Parallel currents attract Antiparallel currents repel

12 s Lecture Question 12.2: Two parallel wires have currents that have the same direction, but differing magnitude. The current in wire A is i; and the current in wire B is 2i. Which one of the following statements concerning this situation is true? (a) Wire A attracts wire B with half the force that wire B attracts wire A. (b) Wire A attracts wire B with twice the force that wire B attracts wire A. (c) Both wires attract each other with the same amount of force. (d) Wire A repels wire B with half the force that wire B attracts wire A. (e) Wire A repels wire B with twice the force that wire B attracts wire A.

13 Similar to Gauss Law, we can find the enclosed current B d s = µ 0 i enc (1) s

14 The right hand rule tells us if the current is positive or negative s Although general, this law is often applied to problems with symmetry

15 Long straight wire: s Let s apply : B d s = B cos(θ)ds = B ds = B(2πr) B(2πr) = µ 0 i enc B = µ 0i 2πr

16 Long thick wire: s Let s apply again: B d s = B cos(θ)ds = B ds = B(2πr) B(2πr) = µ 0 i enc B = µ 0i enc 2πr

17 Long thick wire: s But what is i enc? For uniformly distributed current, we use a ratio: Therefore, i enc = i πr2 πr 2 = i(r/r)2 B = µ 0i 2πR 2 r

18 s Solenoid:

19 Infinite Solenoid: s For this loop, there are four sections b c B d s = B d s + B d s + = a b a = µ 0 i enc = µ 0 Ni = µ 0 (nh)i b B d s = Bh B = µ 0 ni (ideal solenoid) d c B d s + a d B d s

20 Toroid: s This loop is just a simple circle B d s = B(2πr) = µ 0 i enc = µ 0 Ni B = µ 0Ni 1 2π r

21 Lecture Question 12.4: When the switch below is closed, what happens to the portion of the wire that runs inside of the solenoid? s (a) There is no effect on the wire. (b) The wire is pushed downward. (c) The wire is pushed upward. (d) The wire is pushed toward the left. (e) The wire is pushed toward the right.