Download 19 Magnetic Forces and fields

Physics 150 Magne0c forces and fields Chapter 19 Magne0c fields What creates magne0c fields? There is no magne0c charge! Any permanent magnet has two poles. What if we cut a permanent magnet in half? Physics 150, Prof. M. Nikolic 2 Magne0c fields All magnets have at least one north pole (N) and one south pole (S). Physics 150, Prof. M. Nikolic 3 Magne0c field lines Physics 150, Prof. M. Nikolic 4 The Earth’s magne0c field Physics 150, Prof. M. Nikolic 5 Magne0c force on a charged par0cle A charged par0cle moving in a magne0c field experiences a force. !
! !
FB = q v × B
q – electric charge v – par0cle velocity B – magne0c field Vector product?! Magnitude of the magne0c force FB = q vBsin θ
θ is the angle between par0cle velocity and magne0c field Physics 150, Prof. M. Nikolic 6 Magne0c field FB = q vBsin θ
FB
B=
qv
SI unit: Tesla [T] N
N
1 Tesla = 1 T = 1
=1
m
Am
C
s
A 1-­‐T magne0c field is a rather strong magne0c field. 1Tesla= 104 Gauss (G) Physics 150, Prof. M. Nikolic 7 Direc0on of the magne0c force – Right hand rule Physics 150, Prof. M. Nikolic 8 Direc0on of the magne0c force – Right hand rule Physics 150, Prof. M. Nikolic 9 Direc0on of the magne0c force – Right hand rule Physics 150, Prof. M. Nikolic 10 Nota0on To depict a vector oriented perpendicular to the page we use crosses and dots. A cross indicates a vector going into the page (think of the tail feathers of an arrow disappearing into the page). A dot indicates a vector coming out of the page (think of the 0p of an arrow coming at you out of the page). B out of the page Physics 150, Prof. M. Nikolic B into the page 11 Conceptual ques0on – A proton beam Q1
Posi0ve par0cles move upward along the trajectory shown. A magne0c field points toward the right. A. 
B. 
C. 
D. 
Up Down Into the page Out of the page proton beam Physics 150, Prof. M. Nikolic 60 12 Conceptual ques0on – Electron beam Q2
Nega0ve par0cles move upward along the trajectory shown. A magne0c field points toward the right. A. 
B. 
C. 
D. 
Up Down Into the page Out of the page electron beam Physics 150, Prof. M. Nikolic 60 13 Exercise: Moving charge A charge of +23 mC is moving west at 4 m/s. A magne0c field of 30 T is poin0ng straight up. What is the magnitude and direc0on of the force on the charge? How does your answer change if the charge is –23 mC? z !
B
!
v
W What is given: q = 23 mC v = 4 m/s B = 30 T y FB = q vBsin θ
FB = 23⋅10 −3 C ⋅ 4m / s ⋅ 30T sin 90 0
x FB = 2.76 N
Direc0on: into the page or in y direc0on or north If the par0cle was -­‐23 mC the magnitude of the force would be the same FB = 2.76 N but in the opposite direc0on (south, -­‐y direc0on, out of the page). Physics 150, Prof. M. Nikolic 14 What happens if you are given the force but not the magne0c field or the speed? Example: A charge of +3 mC is moving at 3 m/s in a uniform magne0c field of 2 T poin0ng north. The charge experiences an upward magne0c force of 3.6 mN. Find the angle θ? In which direc0on is the par0cles’ velocity? !
F
z W y N FB = q vBsin θ
!
B
!
v
sin θ =
x E FB
q vB
3.6 ⋅10 −3 N
sin θ =
= 0.2
−3
3⋅10 C ⋅ 3m / s ⋅ 2T
θ = sin −1 0.2 = 11.6 0 or 168.4 0
The direc0on of the velocity is either north of east or south of east. Physics 150, Prof. M. Nikolic 15 Mo0on of charges in B field FB = q vBsin θ
FB = q vB
Since v and B are perpendicular By applying second Newton’s law for circular mo0on (Chapter 5) FB = Fcnetripetal = macentripetal
v2
q vB = m
R
The period (the 0me for one full revolu0on) 2π R 2π m
T=
=
v
qB
Physics 150, Prof. M. Nikolic The radius of the circular trajectory mv
R=
qB
16 2π m
T=
qB
Exercise: U – turn A charged par0cle moves into a region of uniform magne0c field, goes through half a circle, and then exits that region. The par0cle is either a proton or an electron (you must decide which). It spends 500 ns within the region. a) Is it an electron or a proton? b) What is the magnitude of B? b) The period (the 0me for 2π m
T=
one full revolu0on) a) It’s a proton qB
But the proton makes a half a revolu0on Thalf
What if we need to find the frequency? 1 qB
f= =
T 2π m
Physics 150, Prof. M. Nikolic πm
=
qB
πm
B=
qThalf
3.14 ⋅1.63×10 −27 kg
B=
500 ×10 −9 s ⋅1.6 ×10 −19 C
B = 0.064 T
17 Crossed E and B fields If a charged par0cle enters a region of space with both electric and magne0c fields present, the force on the par0cle will be !" !!" !!"
!" " !"
F = FE + FB = qE + qv × B
Velocity selector Net force is equal to zero when Physics 150, Prof. M. Nikolic qE = q vB
E
v=
B
18 Applica0on: Mass spectrometer Converts electric poten0al energy into kine0c energy 1
U = −qΔV = mv 2
2
Velocity selector is used to determine speed Es
v=
Bs
Par0cles of different mass will travel different distances before striking the detector. (v, B, Es, Bs, and q can be controlled.) Physics 150, Prof. M. Nikolic 19 Applica0on: Hall effect If an electric current flows through a conductor in a magne0c field, the magne0c field exerts a transverse force on the moving charge carriers which tends to push them to one side of the conductor. Hall Voltage (VH) IB
VH =
ned
n is the density of the charge carriers d is the thickness of the conductor Physics 150, Prof. M. Nikolic 20 Magne0c force on a current carrying wire Current in a wire is a collec0on of moving charges; therefore, a current carrying wire in a magne0c field also experiences a force. Magnitude of the force F = I LBsin θ
L – length of a wire I – current in the wire B – magne0c field θ – angle between magne0c field and current Direc0on of the force – right hand rule Physics 150, Prof. M. Nikolic 21 Conceptual ques0on – Force on a current carrying wire Q3
The diagram shows a straight wire carrying a current into the page. The wire is between the poles of a permanent magnet. The direc0on of the magne0c force exerted on the wire is: A. 
B. 
C. 
D. 
→ ← ↓ ↑ Physics 150, Prof. M. Nikolic 60 22 Exercise: A current carrying wire A 30-­‐cm long conductor is connected to the circuit as shown bellow. The conductor carries current of 140 mA and is placed in magne0c field of 900 mT with the angle between the wire and the field of 30o. a)  What is the direc0on of current? b)  What is the magnitude of the magne0c force? a) Current is always running away from plus (higher poten0al) towards minus (lower poten0al) => current flows to the right F = I LBsin θ
B F = 140 ⋅10 −3 A ⋅0.3m ⋅ 900 ⋅10 −3 T ⋅ sin 30 0
F = 0.019N
Physics 150, Prof. M. Nikolic 23 Force on a current loop F = I LBsin θ
!
Fin
!
Fout
Force due to horizontal segments of the loop is zero because current and field are parallel . Only the ver0cal segments of the loop experience force. Total (net) force Fnet = Fout − Fin
Fnet = IhB − IhB = 0
Physics 150, Prof. M. Nikolic Net magne0c force on a current loop is always equal to zero. 24 Torque on a current loop What about torque? We defined torque in Chapter 8: !
Fin
τ = F r sin θ
!
Fout
r -­‐ distance from axis of rota0on to the force F -­‐ magne0c force on segment θ -­‐ angle between vector r and vector F. Since, there are only two forces ac0ng on the loop w
w
τ = Fout sin θ + Fin sin θ
2
2
Physics 150, Prof. M. Nikolic τ = IhwBsin θ
τ = IABsin θ
A = hw – is the area of the loop 25 Torque on a current loop If the loop has N turns, then τ = NIABsin θ
θ -­‐ angle between the magne0c field and the normal to the area (area vector) The torque will rotate the loop so that the plane of the loop is perpendicular to the magne0c field. Physics 150, Prof. M. Nikolic 26 Conceptual ques0on – Torque Q4
A rectangular loop is placed in a uniform magne0c field with the plane of the loop parallel to the direc0on of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop: A. 
B. 
C. 
D. 
A net torque. A net force. A net force and a net torque. Neither a net force nor a net torque. Physics 150, Prof. M. Nikolic 60 27 Exercise: Torque on a circular coil What is the magnitude of the torque of a circular coil of wire of radius 8 cm carrying a current of 16 mA through 25 loops if the area vector (normal to the area) makes an angle of 60o with respect to an external magne0c field of magnitude 0.2 T? What is given: r = 8 cm I = 16 mA N = 25 θ = 600 T = 0.2 T τ = NIABsin θ
The area of the circular coil A = π r2
A = 3.14 ⋅ (0.08m)2 = 0.02m 2
τ = 25⋅16 ⋅10 −3 A ⋅ 0.02m 2 ⋅ 0.2T ⋅ sin 60
τ = 0.0014Nm
Physics 150, Prof. M. Nikolic 28 Sources of the magne0c field Hans Oersted: Experimental observa0on in 1820: Electric currents can create magne0c fields. Physics 150, Prof. M. Nikolic 29 Direc0on of the magne0c field The Right Hand Rule #2 Physics 150, Prof. M. Nikolic 30 Magne0c field of a long straight wire I µ0 I
B=
2π r
r – distance from the wire µ0 = 4π x 10-­‐7 T·∙m/A –
permeability of free space Examples I I1 out I2 a in Magne0c field a goes into the page at point a Physics 150, Prof. M. Nikolic 31 Exercise: Two long straight wires Consider the long, straight, current-­‐carrying wires shown in the figure. One wire carries a current of 6.2 A in the nega0ve y direc0on; the other carries a current of 4.5 A in the posi0ve x direc0on. Calculate the magnitude and direc0on of the net magne0c field at points A and B. Let’s call a 4.5 A current carrying wire: I1 = 4.5 A è this current produces B1 field And 6.2 A: I2 = 6.2 A è this current produces B2 field B2 B1
At point A: B1 comes out of the page B2 comes into the page Total (net) magne9c field è sum of all fields BA = B1 − B2 =
4π ×10 −7 N / A 2 ⋅ 4.5A 4π ×10 −6 N / A 2 ⋅ 6.2A
BA =
−
2π ⋅ 0.16m
2π ⋅ 0.16m
Physics 150, Prof. M. Nikolic µ 0 I1 µ 0 I 2
−
2π r1 2π r2
BA = 2.13×10 −6 T into the page
32 Exercise: Two long straight wires Consider the long, straight, current-­‐carrying wires shown in the figure. One wire carries a current of 6.2 A in the nega0ve y direc0on; the other carries a current of 4.5 A in the posi0ve x direc0on. Calculate the magnitude and direc0on of the net magne0c field at points A and B. B2 B1
B2 B1
At point B: B1 comes out of the page B2 comes out of the page BB = B1 + B2 =
µ 0 I1 µ 0 I 2
+
2π r1 2π r2
4π ×10 −7 N / A 2 ⋅ 4.5A 4π ×10 −6 N / A 2 ⋅ 6.2A
BB =
+
2π ⋅ 0.16m
2π ⋅ 0.16m
BB = 13.42 ×10 −6 T out of the page
Physics 150, Prof. M. Nikolic 33 Force between two parallel currents Magne0c field of wire a at a loca0on of wire b µ0 I a
Ba =
2π d
Magne0c force on wire b due to the magne0c field Ba Fba = I b LBa
µ 0 LI a I b
Fba =
2π d
Physics 150, Prof. M. Nikolic •  If the currents are parallel, the force is a<rac9ve. •  If the currents are an9-­‐parallel the force is repulsive. L is the length of wire b 34 Exercise: Parallel wires Two parallel wires in a horizontal plane carry currents I1=20 mA and I2=60 mA to the right. The wires each have a length 30 cm and are separated by a distance 4 cm. 1 I1 d
2 I2 What is given: I1=20 mA I2=60 mA L = 30 cm d = 4 cm B1
(a) What are the magnitude and direc0on of the B-­‐field of wire 1 at the loca0on of wire 2? µ 0 I1
B1 =
2πd
4π ×10 −7 N / A 2 ⋅ 0.02A
B1 =
= 10 −7 T
2π ⋅ 0.04m
Physics 150, Prof. M. Nikolic Into the page 35 Exercise: Parallel wires Two parallel wires in a horizontal plane carry currents I1=20 mA and I2=60 mA to the right. The wires each have a length 30 cm and are separated by a distance 4 cm. 1 I1 F21
2 d
I2 What is given: I1=20 mA I2=60 mA L = 30 cm d = 4 cm B1
(b) What are the magnitude and direc0on of the magne0c force on wire 2 due to wire 1? F21 = I 2 LB1
F21 = 0.06A ⋅ 0.3m ⋅10 −7 T = 1.8⋅10 −9 N
Physics 150, Prof. M. Nikolic Up, towards wire 1 – aurac0ve force 36 Magne0c field of a current loop The magne0c field at the center of a current loop: Nµ0 I
B=
2R
N – number of loops of wire (number of turns) R – radius of the loop Physics 150, Prof. M. Nikolic 37 Magne0c field of a solenoid If we stack wire loops next to each other to form a cylinder è we build a solenoid Magne0c field inside a very long solenoid µ 0 NI
B=
L
N – number of turns L – length of the solenoid We can define the number of turns per unit length n: N
n=
L
Direc0on: Right hand rule Physics 150, Prof. M. Nikolic B = µ0nI
38 Ampere’s law Similar to Gauss’s law q
Φ = EA cosθ =
ε0
Ampere’s Law relates the magne0c field on a path to the net current cuvng through the path. circulation = ∑ B||Δl
Ampere’s Law is ∑ B Δl = µ I
||
Physics 150, Prof. M. Nikolic 0 enclosed
39 Exercise: Circula0on A number of wires carry current into or out of the page as indicated. What is the circula0on through the interior of loop 1 and loop 2? ∑ B Δl = µ I
||
0 enclosed
Loop 1: ∑ B Δl = µ (14I − 3I − 6I ) = 5I µ
||
0
0
Loop 2: ∑ B Δl = µ (−14I +16I ) = 2I µ
||
Physics 150, Prof. M. Nikolic 0
0
40 Magne0c materials •  On atomic level – moving electrons (microscopic current loops) create magne0c fields. •  In many materials, these currents are randomly oriented (net magne0c field is zero). •  In some materials, the presence of an external magne0c field can cause the loops to become oriented è Paramagne0sm – orienta0on with an external field è Diamagne0sm – orienta0on against an external field è Ferromagne0sm – line up loops (magne0c domains) without an external field Physics 150, Prof. M. Nikolic 41 Review Magnitude of the magne0c force The period (the 0me for one full revolu0on) FB = q vBsin θ
2π R 2π m
T=
=
v
qB
θ is the angle between par0cle velocity and magne0c field The radius of the circular trajectory Magnitude of the force on a current carrying wire F = I LBsin θ
Physics 150, Prof. M. Nikolic mv
R=
qB
Torque for N turns τ = NIABsin θ
42 Review Magne0c field of a long straight wire Magne0c force between two wires µ 0 LI a I b
Fba =
2π d
µ0 I
B=
2π r
Magne0c field of a solenoid µ 0 NI
B=
L
N
n=
L
B = µ0nI
Ampere’s Law ∑ B Δl = µ I
||
Physics 150, Prof. M. Nikolic 0 enclosed
43