 Phyllis Fleming Physics Physics 108
 Outline - Induction
1. Background

1. Magnetic field lines

1. Magnetic Field Lines for Current in Coil of Wire (Fig. 1a) 1. Curl the fingers of your right hand in the direction of the current. Your thumb points in the direction of the field along the axis of the coil.

2. The field at other points is tangent to the field lines which are dashed in Fig. 1a.

2. Magnetic Field Lines for Bar Magnet (Fig. 1b) 1. Magnetic field lines leave North pole and go to South pole. The field at any point is tangent to the field line.

2. North pole of magnet acts like current carrying coil of wire with current in counterclockwise direction as you view that end of the coil.

3. Before we knew the cause of magnetism was a net flow of charge, we said a circular coil with current in counterclockwise direction as you view that end of the coil acted like a North pole.

4. By experiment we find that a North Pole repels another North pole or a coil with the current in a counterclockwise direction.

2. Magnetic Flux 1. ΦB = B . A = BA cos B,A for B constant.
2. For Fig. 2a, B,A = 90o.  cos 90o = 0  and  ΦB = 0.
3. For Fig. 2b,  ΦB = BA cos B,A.
4. For Fig. 2c, B,A = 0o.  cos 0o = 1  and  ΦB = BA.
5. For Fig. 2d,  notice that 6. If B, A,  or B,A  vary,  ΦB = ∫ B . dA.

3. The Experiment

1. First we calibrate a galvanometer to detect the presence and the direction of a current. In Fig. 3 when current enters the terminal on the right hand side of the galvanometer, the needle deflects to the left. 2. In Fig. 4a, the bar magnet is at rest. There is no current in coil of wire and the galvanometer does not deflect. 3. In Fig. 4b, the magnet moves toward the coil and the galvanometer deflects toward the left indicating an induced current and associated induced magnetic field (shown by dashed line) opposing the field of the magnet. 4. In Fig. 4c, the rate of change of the magnetic flux produced by the magnet has increased and there is a greater current in the galvanometer. 5. In Fig. 4d, the magnet is inside the coil at rest. Current is zero. 6. In Fig. 4e, the magnet moves to the right. Current in coil reverses. 4. Results of the Experiment

The induced current creates an induced magnetic field that opposes the change produced by the external magnetic field. In this experiment the bar magnet provided the external field.

1. When the magnet is at rest there is no change of flux or induced current (Fig. 4a above).

2. When the North pole of the magnet moved toward the right face of the coil, the induced current in the coil was counterclockwise as viewed from its right end of the coil. Another way of looking at this is the right end of the coil acts like a North pole and repels the North pole of the magnet as you move the magnet toward it. You must exert a force through a distance doing work to induce a current in the coil (Fig. 4b above).

3. The greater rate of change of magnetic flux, the greater induced current. As the magnet gets nearer to the coil, the flux changes more rapidly. You can also see this by bringing up two magnets at the same speed of one or bringing up one more rapidly (Fig. 4c above).

4. When the magnet is at rest, there is no change of flux and no current (Fig. 4d above).

5. When the North pole is moved away from the right end of the coil the current in the coil reverses. The end of the coil nearest the magnet acts like a South pole and you must do work to remove the North pole of the magnet and induce a current in the coil (Fig. 4e above).

1. Induced emf = the rate of change of magnetic flux.

1. Since ΦB = ∫ B . dA, you can change the magnetic flux by changing the magnetic field, the area through which the field exists or the angle between the magnetic field and the area.

2. The direction of the induced current is given by Lenz's Law: The direction of the induced current is such to oppose the change that produced it. Lenz's Law is really a statement of conservation of energy. In order to get an induced emf or current, you must do work.

2. Motional emf 1. As conductor in Fig. 5 moves to the right, the electrons in the conductor move with its velocity v and experience a magnetic force down equal to qvB, where q is the charge of the electron. They accumulate at the bottom of the conductor leaving the top of the conductor with an excess of positive charge.

2. This separation of charge produces an electric field from a to b that results in an electric force up on the electron equal to qE, where q is the charge of the electron.

3. When the electric force equals the magnetic force,  qE = qvB,  the motion of the electron ceases and the electric field in the conductor E = vB. The potential difference between the ends of the conductor
Vab = E = (Bv) 4. The motional emf = Bv .  In Fig. 5,  B was perpendicular to v.
If the magnetic field is not perpendicular to the velocity, the motional emf = Bv sin B,v.

3. Alternating Current Generator

1. In Fig. 2b above, the magnetic flux ΦB = BA cos B,A.

2. Let B,A = Θ =  ωt, where ω equals the angular velocity of the rotating coil.  Then ΦB = BA cos ωt  and  ε = - NdΦB/dt
= N ωBA sin ωt,  where N = the number of turns of the coil.

3. Figure 6a and 6b below are plots of the magnetic flux and the electromotive force as a function of time t, respectively. 4. Sample problems in 108 Problem Set for Induction:  1-10.

3. The Production of an Electric Field by a Changing Magnetic Flux

1. The electromotive force ε equals the energy per unit charge and the electric field E equals the force per unit charge. In Fig. 7 above we can explain the motion of charge in the wire as an induced current. We can also explain it as the production of an electric field that, in turn, produces a motion of positive charge as a current. The work done per unit charge around the loop of length 2 πr equals ε = E(2 πr) = - dΦ/dt.

2. In general we may write E • d = B • dA

3. Sample problem in 108 Problem Set for Induction:  12

4. Self-inductance L = NΦB/I

1. Calculation of L for a solenoid

For a solenoid,  B = µonI = µo(N/ )I,
where n = the number of turns per unit length = N/ .

B = µo(N2/ )I πr2,  where r is the radius of the solenoid.

L = NΦB /I = µo(N2/ )( πr2).

2. Energy in Magnetic Field

1. Since L = NΦB/I,  L = N (dΦB/dt)/(dI/dt) = ε/(dI/dt).

2. Energy = dU = (power)dt = εI dt = L(dI/dt)I dt = LI dI.
U = Lo I I dI = 1/2 LI2.

3. Using L from solenoid, the magnetic energy
UB = 1/2 [(µo(N2/ )( πr2)]I2
= 1/2[µo(N2/ )( πr2)][B/µo(N/ )]2
= (1/2µ0) B2[ πr2]
= (1/2µo) B2[volume]

4. While we have derived this from L for a solenoid, in general
UB = 1/2 (B2o)A(length)
and the magnetic energy density
uB = (1/2µo )B2.
Compare with electric energy density
uE = (1/2) εo E2.

5. Example

Find the magnetic energy stored for a < r < b for the coaxial inductor shown in Fig. 8 below.

Assume that the inductor is sufficiently long ( >> b), or that the gap is sufficiently small, that end effects can be neglected. From Ampere's law, we know that for  a < r < b,   B = µoI/2 πr. The volume element is the volume of a cylindrical shell with radius r, surface area 2 πr , and thickness dr.

dV = 2 πr dr

U = (1/2µo) ab B2 dV
= (1/2µo) ab oI/2 πr] 2 2 πr dr
= [µoI2 /4 π] ln (b/a)

3. Sample problems in 108 Problem Set for Induction:  11, 15.

5. R-L Circuits

1. Inductance  L = ΔVL/(dI/dt),  where ΔVis the potential difference across the inductor and dI/dt is the rate of change of current in the inductor.

1. Unit of Inductance = 1 Henry (H) = 1V-s/A

2. Similar to resistance and capacitance, the inductance depends only on the geometry of the conductor.

2. Build up of current 1. When a source of emf is connected across a resistor and inductor wired in series with a battery (Fig. 9b above), the current goes from zero to some finite value in a very short time. Since the rate of change of current is very high, the potential difference across the inductor = Vcb= L(dI/dt) is very high, while the small current makes the potential difference across the resistor Vac = IR very small. As time passes, the current builds up and the rate of change of current decreases. After a very long time, it is as though the inductor were not in the circuit and the total emf is across the resistor giving a maximum current = ε/R. A plot of Vac and Vcb (with dashes) for the build up of the current is shown in Fig. 9a above.

2. For the build up of the current,
I(t) = ( ε/R)(1 - e-Rt/L)
Vac(t) = IR = ε(1 - e-Rt/L)
Vcb(t) = L dI/dt = L(-  ε/R) e-Rt/L (-R/L) = ε e-Rt/L
3. Decay of the Current

1. When the battery is removed and a wire is placed across the resistor and the inductor (Fig. 9b above), the potential difference Vab = 0. As the current decreases, the inductor's "back emf" opposes the decrease in current. A plot of Vac and Vbc (with dashes) for the decay of the current is shown in Fig. 9a’ above.

2. For the decay of the current,
I(t) = ( ε/R)e-Rt/L
Vac(t) = IR = ε e-Rt/L
Vcb(t) = L dI/dt = L( ε/R)(e-Rt/L)(-R/L) = - εe-Rt/L
4. Sample problem in 108 Problem Set for Induction:  13

6. L-C Circuits

1. Comparison of LC-oscillator with spring Spring Oscillator Fnet = ma Vab = Vac + Vcb - kx = m dv/dt 0 = q/C + L dI/dt Since v = dx/dt, Since I = dq/dt, dv/dt = d/dt(dx/dt) = d2x/dt2 dI/dt = d/dt(dq/dt) = d2q/dt2 -kx = m d2x/dt2   or -q/C = L d2q/dt2   or d2x/dt2 + (k/m)x = 0 d2q/dt2 + (1/LC)q = 0 ω = 2 πf = (k/m)1/2 ω = 2 πf = (1/LC)1/2

2. Charge, Current, and Potential Energy as a function of time for L-C circuit

1. For fully charged capacitor at time t = 0,
q(t) = Q0 cos ωt, with ω = 2 πf = (1/LC)1/2.

2. I(t) = dq/dt = I(t) = ωQ0 sin ωt

3. Electric potential energy Ue = 1/2 q2/C = (1/2C) Q02 cos2 ωt

4. Magnetic potential energy Um = 1/2 LI2 = (L/2) ω2Q02 sin2 ωt

5. Since ω= 2 π/T,
maximum electric energy occurs at  t = 0, T/2, 2T . . . maximum magnetic energy occurs at  t = T/4, 3T/4 . . .
Sample problem in 108 Problem Set for Induction:  14

7. Electromagnetic Waves

1. From Faraday's law we find (Equation 1)

This equation states that a changing magnetic flux produces an electric field.

2. From Ampere's law (Equation 2)

This equation states that a current produces a magnetic field.

1. Since nature is amazingly symmetrical, Maxwell believed that a changing electric flux should produce a magnetic field.

2. As an example, he pointed out the discontinuity in the magnetic field while a capacitor is charged. In Fig. 11 below, you see the magnetic field lines associated with the current entering the capacitor and leaving the capacitor, but there is apparently no magnetic field inside the capacitor. 3. While the capacitor is charging, the electric field inside the capacitor is changing since E = q/ εoA.

dE/dt = (dq/dt)/ εoA   or   (dq/dt) = current = ( εoA) dE/dt.

Maxwell called this current a displacement current Id and generalized it by writing Id = εo d/dt ∫ E . dA         (Equation 3)

4. With this addition we rewrite Eq. 2 (Equation 4)
Eq. 4 states that a changing electric flux produces a magnetic field even in the absence of a current.

5. Maxwell suggested that a changing electric field produced a magnetic field, which if changing, produced an electric field, which if changing, produced a magnetic field, etc. He hypothesized that these disturbances would be transmitted as an electromagnetic wave. Figure 12a below is a representation of an electromagnetic wave traveling in the X-direction with the electric vector in the X-Y plane and the magnetic vector in the Y-Z plane. 6. The equations of motion for the E and B vectors are
E = Ep cos (kx - ωt)              (Equation 5)
B = Bp cos (kx - ωt)            (Equation 6)
where k = 2 π/λ and ω = 2 πf.

Since λf = c,  ω/k = (2 πf)/(2 π/λ) = λf = c.

7. Evaluating paths in Fig. 12b and 12c. Let us evaluate Eq.1 for the path in Fig. 12b above.

You choose the direction of going around the loop by pointing the thumb of your right hand in the direction of B. Your fingers then curl clockwise in the direction of the path. For this path, the E, ds along the right-hand side of the loop is 0o and along the left-hand side of the loop it is 180o. E ds = (E + dE) ds on the right side of length h and –E ds for the left side. For the sides of length dx, E, ds is 90o and cos 90o = 0. There is no contribution for these lengths.

Letting B = the average value of the loop of area h dx,
-d/dt ∫ B . dA = -dB/dt hdx.

Thus becomes
(E + dE)h – Eh = - dB/dt h dx   or   dE/dx = - dB/dt,

or in terms of partial derivatives,

∂E / ∂x = - ∂B / ∂t          (Equation 7)

8. Write Maxwell's equation in free space, that is, assume there is
no "real" current I.

1. Then (Equation 4)
becomes (Equation 4’)

2. Evaluate Eq. 4’ for the path of Fig. 12c above.

You choose the direction of going around the loop by pointing the thumb of your right hand in the direction of E. Your fingers then curl in the direction of the path. For this path, the B, ds along the left-hand side of the loop is 0o and along the right-hand side of the loop it is 180o.

B . ds = - (B + dB)dx on the right side of length h and B dx for the left side. For the sides of length dx, the B, ds = 90o and there is no contribution to those sides. =-dB dx

Letting E = the average value of the loop of area h dx,
µo εo d/dt ∫ E . dA = µo εo dE/dt hdx  or  Eq. 4' becomes
- dB/dx = µo εo dE/dt
or in terms of partial derivatives
- ∂B / ∂x = µo εo ∂E / ∂t         (Equation 8)
Taking the partial derivative of Eq. 7 with respect to x gives
2E / ∂x = - ( ∂ / ∂x)∂B / ∂t         (Equation 9)
Taking the paratial derivative of Eq. 8 with respect to t gives
- (∂ / ∂t)∂B / ∂x = µo εo (∂2E / ∂t2)        (Equation 10)
Since the order of differentiation does not make any difference, you can equate the left side of Eq. 9 to the right side of Eq. 10:
∂2E / ∂x2 = µo εo (∂2E / ∂t2)         (Equation 11)
Compare Eq. 11 with wave equation of wave along rope
2y / ∂x2 = (1 / v2) (∂2y / ∂t2)
The equations are identical for a wave traveling with a velocity (1/µo εo)1/2 and changing electric field E, and a wave with a velocity v with changing displacement of the rope y from the equilibrium position, if c = (1/µo εo)1/2.

3. Since µo= 4 π x 10-7 N/A2 and εo = 8.85 x 10-12 C2/N-m2, (1/µo εo)1/2 = 1/[(4 π x 10-7 N/A2)( 8.85 x 10-12 C2/N-m2)]1/2
= 3.00 x 108 m/s!

As the late Richard Feynman said in a rewrite of Genesis,
"Let there be electricity and magnetism and there will be light."

c = (1/µo εo)1/2

9. If you take the derivative of Eq. 5 with respect to x and the derivative of Eq. 6 with respect to t and substitute into Eq. 7, you find
- kEp sin(kx - ωt) = - ωBp sin(kx - ωt)   or
Ep = (ω/k)Bp = {2 πf/(2 π/ λ)Bp   or
Ep = Bp(f λ)
Bp = Ep/c

4. The Poynting vector S

1. S = 1/µo E x B.

2. S is the energy/second per m2.  Since energy/second = power = P,
S = power/area.

3. Frequently you can solve problems by checking units.
The unit of power is J/s = W.
The unit of S is W/m2.

4. The direction of S is given by E x B.
In Fig. 12,  E = Ej  and  B = Bk.
Since j x = i, and S = 1/µo E x B,  S is in the positive X-direction.

5. Magnitude of S = EB/µo = E2/cµo.

6. Average value of Sav = (E2)av/cµo = (Ep)2[cos2(kx - ωt)]av/cµo. [cos2(kx - ωt)]av = 1/2  and  Sav = (Ep)2/2cµo.

7. Power (W) = Energy/time (J/s).

8. Average value of S = Intensity (W/m2) = Power (W)/Area (m2).

5. Sample problems in 108 Problem Set for Induction:  16-20.

8. Momentum and Pressure of an Electromagnetic Wave

1. Assume that an initially at rest, positive charge q is in the path of a wave (Fig. 13 below). The force Fe on it by the electric field is qE. In time Δt, the electric field moves the charge a distance v Δt and does work Fv Δt = (qE) v Δt. The charge's increase in energy = ΔU = qE v Δt  or  qvE =  ΔU/ Δt. 2. Once the charge is in motion, the magnetic field exerts a force on the charge to the right. Although the magnetic force does no work, it does give the charge momentum.

The magnetic force Fm = qvB = qvE/c, since B = E/c for an electromagnetic wave.

Thus Fm = qvB = qvE/c = ΔU/c Δt   or   Fm Δt = ΔU/c.

Since F = Δmv/ Δt = Δp,   Δp = F Δt = ΔU/c.  If a particle absorbs energy ΔU, it acquires a momentum ΔU/c. The momentum of the particle must have belonged to the wave!

3. Pressure = P = Force/Area.
For a wave, since F Δt = ΔU/c,
F = [( ΔU/ Δt)/c]  and
P = [( ΔU/ Δt)]/c)Area = [( ΔU/ Δt)/area]/c.
Power = [( ΔU/ Δt)].
(Power)/Area = Sav.
Thus P = Sav/c.

9. Alternating Currents

1. The basic problem is an analysis of alternating current (ac) to find expressions for the current amplitude Ip and the phase angle Φ where the current I(t) = Ip sin (ωt + Φ), when V(t) = Vp sin ωt is applied to a circuit.

2. Single Circuit Elements

1. Resistor R only (Fig. 14a) 1. V(t) = Vp sin ωt = VR = IR R.
IR = Vp sin ωt/R.
IR = IpR sin ωt, where IpR = VpR/R.

The current and the potential difference across R are in phase (Fig. 14c).

2. The phasor diagram is shown in Fig. 14c for a resistive circuit only. The maximum values for the current in R and the potential difference across R are IpR and VpR, respectively. The instantaneous values of the current and the potential difference across R, IR and VR, respectively, are shown for the time indicated in Fig. 14b.

Notice that IR = IpR sin ωt and VR = VpR sin ωt .

2. Capacitor C only (Fig. 15a) 1. V(t) = Vp sin ωt = VC = q/C.  q = CVC = CVpCsin ωt.
IC = dq/dt = ωCVpC cos ωt = ωCVpC sin (ωt + π/2) because
(sin ωt cos π/2 + cos ωt sin π/2) = cos πt.
IC = Ipc sin (ωt + π/2), where Ipc = VpC/(1/ωC).
1/ωC = XC.

The capacitive reactance takes the place of R in a total resistive circuit. The current in the capacitive circuit leads the potential difference across the capacitor (Fig. 15b).

2. The phasor diagram for a capacitor only circuit is shown in Fig. 15c. The maximum values for the current for C and the potential difference across C are IpC and VpC, respectively. The instantaneous values of the current and the potential IC and VC are shown at the time indicated in Fig. 15b.

Notice that IC = IpC sin (ωt + π/2).   VC = VpC sin ωt.

3. Inductor only (Fig. 16a) 1. V(t) = VpL sin ωt = VL = L dIL/dt.

Separating variables,

VpL sin ωt dt = L dIL
(VpL / L) sin ωt dt = dIL
- (VpL / ωL) [cos ωt - 1] = IL - IoL
IoL = - (VpL / ωL)
IL (t) = - (VpL / ωL) cos ωt
= (VpL / ωL) sin (ωt - π/2)
(VpL/ωL) cos ωt = (VpL/ωL) sin (ωt - π/2) because
(sin ωt cos π/2 - cos ωt sin π/2) = - cos ωt.

IL = IpL sin (ωt - π/2), where I pL = VpL/(ωL).
ωL = the inductive reactance = XL.

The current for the inductive circuit lags the potential difference across the inductance by π/2 as shown in Fig. 16b.

2. The phasor diagram for an inductor only circuit is shown in Fig. 16c. The maximum values for the current in L and the potential difference across are IpL and VpL, respectively. The instantaneous values of the current and the potential IL and VL, respectively, are shown at the time indicated in Fig. 16b.

Notice in the phasor diagram IL = IpL sin(ωt - π/2).
Vc = VpC sin ωt.

10. RLC Series Circuit   1. The applied potential difference across the circuit V(t) = Vp sin ωt (Fig. 17a) and the current I(t) = Ip sin (ωt - Φ) is the same everywhere in the series circuit.

The potential difference across the resistor = VR(t) = IR = [Isin (ωt - Φ)]R is in phase with the current (Fig. 17b and Fig. 18b).

The potential difference across the inductor = VL(t) leads the current by π/2 (Fig. 17c and Fig. 18b). VL(t) = [Ip sin (ωt - Φ + π/2)]ωL.

The potential difference across the capacitor VC(t) lags the current by π/2 (Fig. 17d and Fig. 18b). VC(t) = [Ip sin (ωt - Φ - π/2)](1/ωC)

2. In Fig. 18c,
Vp2 = VpR2 + (VpL – VpC)2
Vp2 = (IpR}2 + (IpωL – Ip/ωC)2
Ip = Vp/[R2 +(ωL –1/ωC)]1/2 = Vp/Z
tan Φ = (ωL –1/ωC)]/R
3. Phasor Diagrams Revisited 1. While the previous phasor diagrams allow you to take projections and find the voltage or current at any time, you can simplify the diagrams by drawing them as shown in Fig. 19 b and c above.

2. Draw R and I along the +X-axis, ωL along the +Y-axis, and 1/ωC along the –Y axis (Fig. 19b)

3. Since I is the same in the resistance, inductance, and capacitance
(Fig. 19c),

1. VpR is drawn along the +X-axis because VR is in phase with I.
2. VpL is drawn along the +Y-axis because VL leads I by π/2.
3. VpC is drawn along the -Y-axis because VC lags I by π/2.

4. From Fig. 19b,  tan Φ= (ωL – 1/ωC)/R.

5. From Fig. 19c,
Vp = [VpR2 + (VpL2 – VpC2)]1/2
Vp = {(IpR)2 + [(IpωL)2 – (Ip/ωC)2]}1/2   or
Ip = Vp/[R2 + (ωL – 1/ωC)2]1/2

4. For the circuit of Fig. 19a,

Vp = 10 V, ω = 120 πs-1, R = 4.0 ω, L = 0.0159 H, and C = 0.884 mF.
XL = ωL = 6.0 ω
XC = 1/ωL = 3.0 ω
Ip = Vp/[R2 + (ωL – 1/ωC)2]1/2 = 10 V/[16 + (6 – 3)2]1/2 ω= 2.0 A
VpR = IpR = 8.0 V
VpL = Ip (XL) = 12 V
VpLc= Ip (Xc) = 6 V

5. The maximum value of Ip occurs for ωL = 1/ωC, or the resonant frequency ωo = 1/(LC)1/2.  At resonance, Ip = Vp/[R2 + (ωL – 1/ωC)2]1/2 = Vp/R. The resonance curve for the circuit of Fig. 19a is shown in Fig. 20 below. 6. Power = IV
= [Ip sin (ωt - Φ)][Vp sin ωt}
= IpVp [sin ωt cos Φ - cos ωt sin Φ)[sin ωt]
= IpVp [sin2 ωt - cos ωt sin ωt]
= IpVp [sin2 ωt cos Φ – 1/2(sin 2ωt](sin Φ)
The average value of sin2 ωt is 1/2 and the average value of sin 2ωt is 0. Pav = (IpVp /2) cos Φ.  Because cos Φ = R/[R2 + (ωL – 1/ωC)2]1/2,
Pav = (Ip2/2)R = I2rms R, where the root-mean-square value of the current is defined as Ip/(2)1/2 and the root-mean-square value of the voltage is defined as Vp/(2)1/2.

7. Sample problems in 108 Problem Set for Induction:  22-26.

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