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.
      
      

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

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.
   
   
   
   

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. Since Φ_B = ∫ 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.
   
   

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

    V_ab = 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. 
   
   

1. In Fig. 2b above, the magnetic flux Φ_B = BA cos B,A.
   
   
2. Let B,A = Θ =  ωt, where ω equals the angular velocity of the rotating coil.  Then Φ_B = BA cos ωt  and  ε = - NdΦ_B/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.
   
   
   
   
   
   

1. Since L = NΦ_B/I,  L = N (dΦ_B/dt)/(dI/dt) = ε/(dI/dt).
   
   
2. Energy = dU = (power)dt = εI dt = L(dI/dt)I dt = LI dI.
   U = L_o ∫^I I dI = 1/2 LI².
   
   
3. Using L from solenoid, the magnetic energy

   U_B = 1/2 [(µ_o(N²/)( πr²)]I²
        = 1/2[µ_o(N²/)( πr²)][B/µ_o(N/)]²
        = (1/2µ₀) B²[ πr²]
        = (1/2µ_o) B²[volume]

   
   
4. While we have derived this from L for a solenoid, in general

   U_B = 1/2 (B²/µ_o)A(length)

   and the magnetic energy density

   u_B = (1/2µ_o )B².

   Compare with electric energy density

   u_E = (1/2) ε_o 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 = µ_oI/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 = (1/2µ_o) _a∫^b B² dV
      = (1/2µ_o) _a∫^b [µ_oI/2 πr] ² 2 πr dr
      = [µ_oI²/4 π] ln (b/a)
   
   
   

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.
   
   

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 = V_cb= L(dI/dt) is very high, while the small current makes the potential difference across the resistor V_ac = 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 V_ac and V_cb (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)
   V_ac(t) = IR = ε(1 - e^-Rt/L)
   V_cb(t) = L dI/dt = L(-  ε/R) e^-Rt/L (-R/L) = ε e^-Rt/L

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
   V_ac(t) = IR = ε e^-Rt/L
   V_cb(t) = L dI/dt = L( ε/R)(e^-Rt/L)(-R/L) = - εe^-Rt/L

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 U_e = 1/2 q²/C = (1/2C) Q₀² cos² ωt
   
   
4. Magnetic potential energy U_m = 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
   
   
   

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/ ε_oA.
   
   dE/dt = (dq/dt)/ ε_oA   or   (dq/dt) = current = ( ε_oA) dE/dt.
   
   Maxwell called this current a displacement current I_d and generalized it by writing I_d = ε_o 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 = E_p cos (kx - ωt)              (Equation 5)
   B = B_p 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 0^o and along the left-hand side of the loop it is 180^o.
   
   
   
   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 90^o and cos 90^o = 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 0^o and along the right-hand side of the loop it is 180^o.
      
      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 = 90^o and there is no contribution to those sides.
      
      =-dB dx
      
      Letting E = the average value of the loop of area h dx,
      µ_o ε_o d/dt ∫ E ^. dA = µ_o ε_o dE/dt hdx  or  Eq. 4' becomes

      - dB/dx = µ_o ε_o dE/dt

      or in terms of partial derivatives

      - ∂B / ∂x = µ_o ε_o ∂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 = µ_o ε_o (∂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 = µ_o ε_o (∂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/µ_o ε_o)^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/µ_o ε_o)^1/2.
      
      
   3. Since µ_o= 4 π x 10^-7 N/A² and ε_o = 8.85 x 10^-12 C²/N-m², (1/µ_o ε_o)^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/µ_o ε_o)^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

   - kE_p sin(kx - ωt) = - ωB_p sin(kx - ωt)   or
       E_p = (ω/k)B_p = {2 πf/(2 π/ λ)B_p   or
       E_p = B_p(f λ)
       B_p = E_p/c

   
   

1. S = 1/µ_o 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 k = i, and S = 1/µ_o E x B,  S is in the positive X-direction.
   
   
5. Magnitude of S = EB/µ_o = E²/cµ_o.
   
   
6. Average value of S_av = (E²)_av/cµ_o = (E_p)²[cos²(kx - ωt)]_av/cµ_o. [cos²(kx - ωt)]_av = 1/2  and  S_av = (E_p)²/2cµ_o.
   
   
7. Power (W) = Energy/time (J/s).
   
   
8. Average value of S = Intensity (W/m²) = Power (W)/Area (m²).
   
   

F = [( ΔU/ Δt)/c]  and
P = [( ΔU/ Δt)]/c)Area = [( ΔU/ Δt)/area]/c.

1. Resistor R only (Fig. 14a)
   
   
   
   
   1. V(t) = V_p sin ωt = V_R = I_R R.
      I_R = V_p sin ωt/R.
      I_R = I_pR sin ωt, where I_pR = V_pR/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 I_pR and V_pR, respectively. The instantaneous values of the current and the potential difference across R, I_R and V_R, respectively, are shown for the time indicated in Fig. 14b.
      
      Notice that I_R = I_pR sin ωt and V_R = V_pR sin ωt .
      
      
      
2. Capacitor C only (Fig. 15a)
   
   
   
   
   1. V(t) = V_p sin ωt = V_C = q/C.  q = CV_C = CV_pCsin ωt.
      I_C = dq/dt = ωCV_pC cos ωt = ωCV_pC sin (ωt + π/2) because
      (sin ωt cos π/2 + cos ωt sin π/2) = cos πt.
      I_C = I_pc sin (ωt + π/2), where I_pc = V_pC/(1/ωC).
      1/ωC = X_C.
      
      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 I_pC and V_pC, respectively. The instantaneous values of the current and the potential I_C and V_C are shown at the time indicated in Fig. 15b.
      
      Notice that I_C = I_pC sin (ωt + π/2).   V_C = V_pC sin ωt.
      
      
      
3. Inductor only (Fig. 16a)
   
   
   
   
   1. V(t) = V_pL sin ωt = V_L = L dI_L/dt.
      
      Separating variables,
      
      V_pL sin ωt dt = L dI_L
      (VpL / L) sin ωt dt = dI_L
      - (V_pL / ωL) [cos ωt - 1] = I_L - I_oL
      I_oL = - (V_pL / ωL)
      I_L (t) = - (V_pL / ωL) cos ωt
               = (V_pL / ωL) sin (ωt - π/2)
      (V_pL/ωL) cos ωt = (V_pL/ωL) sin (ωt - π/2) because
      (sin ωt cos π/2 - cos ωt sin π/2) = - cos ωt.
      
      I_L = I_pL sin (ωt - π/2), where I _pL = V_pL/(ωL).
      ωL = the inductive reactance = X_L.
      
      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 I_pL and V_pL, respectively. The instantaneous values of the current and the potential I_L and V_L, respectively, are shown at the time indicated in Fig. 16b.
      
      Notice in the phasor diagram I_L = I_pL sin(ωt - π/2).
      V_c = V_pC sin ωt.
      
      
      

V_p² = V_pR² + (V_pL – V_pC)²
V_p² = (I_pR}² + (I_pωL – I_p/ωC)²
I_p = V_p/[R² +(ωL –1/ωC)]^1/2 = V_p/Z
tan Φ = (ωL –1/ωC)]/R

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. V_pR is drawn along the +X-axis because V_R is in phase with I.
   2. V_pL is drawn along the +Y-axis because V_L leads I by π/2.
   3. V_pC is drawn along the -Y-axis because V_C lags I by π/2.
      
      
4. From Fig. 19b,  tan Φ= (ωL – 1/ωC)/R.
   
   
5. From Fig. 19c,

   V_p = [V_pR² + (V_pL² – V_pC²)]^1/2
   V_p = {(I_pR)² + [(I_pωL)² – (I_p/ωC)²]}^1/2   or
   I_p = V_p/[R² + (ωL – 1/ωC)²]^1/2

   
   
   

= [I_p sin (ωt - Φ)][V_p sin ωt}
= I_pV_p [sin ωt cos Φ - cos ωt sin Φ)[sin ωt]
= I_pV_p [sin² ωt - cos ωt sin ωt]
= I_pV_p [sin² ωt cos Φ – 1/2(sin 2ωt](sin Φ)