Electromagnetic Induction

Setup: An inductor of self inductance $L$ is connected to a source. The current grows from $0$ to its final value $I_0$. While the current grows, a back emf $\varepsilon = -L\dfrac{dI}{dt}$ opposes it, and the source must do work against this back emf. That work is stored as energy in the magnetic field of the inductor.

1. When the current is $I$, the power supplied against the back emf is

$$P = |\varepsilon|\,I = L\frac{dI}{dt}\,I$$

2. The work done in a small time $dt$ is

$$dW = P\,dt = LI\,dI$$

3. The total work done as the current grows from $0$ to $I_0$ is

$$W = \int_0^{I_0} LI\,dI = L\left[\frac{I^2}{2}\right]_0^{I_0}$$ $$\boxed{U = \tfrac{1}{2}LI_0^2}$$

This is the energy stored in the magnetic field of the inductor.

(a) Magnetic flux

The magnetic flux through a surface is the total number of magnetic field lines passing normally through it. For a uniform field $B$ making angle $\theta$ with the normal to an area $A$,

$$\phi = \vec B\cdot\vec A = BA\cos\theta$$

Its SI unit is the weber (Wb), where 1 Wb = 1 T m$^2$.

(b) Faraday's laws of electromagnetic induction

1. First law: Whenever the magnetic flux linked with a circuit changes, an emf is induced in it. The emf lasts only as long as the flux keeps changing.
2. Second law: The induced emf equals the rate of change of flux linkage:

$$\varepsilon = -N\frac{d\phi}{dt}$$

(c) Numerical

Given: $N = 5000$, $\Delta\phi = 8\times10^{-4}$ Wb, $\Delta t = 0.1$ s

$$|\varepsilon| = N\frac{\Delta\phi}{\Delta t} = 5000\times\frac{8\times10^{-4}}{0.1} = 5000\times8\times10^{-3}$$ $$\boxed{|\varepsilon| = 40\ \text{V}}$$

(d) Aluminium ring and moving magnet

      ////////////
           |
           |  thread
          ( )  ring
  N[====]S
   <----  magnet moved away

• As the magnet (S pole facing the ring) moves away, the magnetic flux through the ring decreases. An emf and a current are induced in the ring.
• By Lenz's law the induced current opposes this decrease. So the face of the ring nearest the magnet becomes an N pole, which attracts the receding S pole. The ring is pulled after the magnet, so the ring follows the magnet.
• If the magnet is moved towards the ring, the flux through the ring increases. The near face of the ring then becomes an S pole, which repels the approaching S pole, so the ring moves away from the magnet (it is repelled).

Aluminium is light and a good conductor, so the induced current is large and the effect is easy to see.

(b) $\dfrac{1}{60}$ Volt

$\dfrac{d\phi}{dt} = \dfrac{1\ \text{Wb}}{60\ \text{s}} = \dfrac{1}{60}$ Wb s$^{-1}$. Since $|\varepsilon| = \dfrac{d\phi}{dt}$ and 1 Wb s$^{-1}$ = 1 V, the induced emf is $\dfrac{1}{60}$ V.

a) Self inductance

Self inductance $L$ of a coil is the magnetic flux linkage with the coil per unit current flowing through it, $N\phi = LI$. It also equals the emf induced in the coil when the current in it changes at the rate of 1 A s$^{-1}$, since $\varepsilon = -L\dfrac{dI}{dt}$. Its SI unit is the henry (H).

b) Direction of induced current in the loop

           .-----.
          /       \      loop (field of wire here: out of page)
          \       /
  A --------'---'---------> B    current I increasing

• By the right-hand grip rule, a current flowing from A to B (to the right) produces a magnetic field out of the page in the region above the wire, where the loop lies.
• As the current increases, the outward flux through the loop increases.
• By Lenz's law, the induced current must oppose this increase. So it produces a field into the page inside the loop.
• A current that gives a field into the page is clockwise (seen from the front).

So the induced current in the loop flows clockwise, and it lasts only while the current in AB keeps increasing.

c) Numerical

Given: $N = 100$, $I = 2$ A, $\phi = 5\times10^{-3}$ Wb

$$N\phi = LI \;\Rightarrow\; L = \frac{N\phi}{I} = \frac{100\times5\times10^{-3}}{2}$$ $$\boxed{L = 0.25\ \text{H}}$$

Faraday's laws

1. Whenever the magnetic flux linked with a circuit changes, an emf is induced in it. The emf lasts only as long as the flux keeps changing.
2. The magnitude of the induced emf equals the rate of change of flux linkage: $\varepsilon = -N\dfrac{d\phi}{dt}$.

Numerical

Given: $\phi = 4t^2 + 5t + 2$ (Wb), $t = 3$ s, and a single-turn coil ($N = 1$).

$$\frac{d\phi}{dt} = 8t + 5$$

At $t = 3$ s:

$$\frac{d\phi}{dt} = 8(3) + 5 = 29\ \text{Wb s}^{-1}$$ $$\boxed{|\varepsilon| = \left|\frac{d\phi}{dt}\right| = 29\ \text{V}}$$

(B) 10 Weber

The coil is perpendicular to the field, so $\theta = 0^\circ$ and $\phi = BA = 10^{3}\times10^{-2} = 10$ Wb.

a) Lenz's law

The direction of the induced emf (or induced current) is always such that it opposes the change in magnetic flux that produces it.

b) Graph of induced emf for a coil rotating in a uniform field

For a coil rotating with angular velocity $\omega$, $\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$, so the graph is a sine curve:

 emf
  ^
e0|         *******
  |      ***       ***
  |   ***             ***
  | **                   **
 0+*-----------------------*-----------------------*--> t
  |                         **                   **
  |                           ***             ***
  |                              ***       ***
-e0                                 *******
  0           T/4         T/2         3T/4        T

The emf is zero when the plane of the coil is perpendicular to the field. It is largest ($\pm\varepsilon_0$) when the plane of the coil is parallel to the field.

c) Energy stored in an inductor

Setup: An inductor of inductance $L$ is connected to a source. The current grows from $0$ to $I_0$. The back emf $\varepsilon = -L\dfrac{dI}{dt}$ opposes the growth, and the work the source does against it is stored in the magnetic field.

1. At an instant when the current is $I$, the rate at which work is done against the back emf is

$$P = |\varepsilon|\,I = LI\frac{dI}{dt}$$

2. The work done in time $dt$ is

$$dW = P\,dt = LI\,dI$$

3. Integrate from $I = 0$ to $I = I_0$:

$$W = \int_0^{I_0} LI\,dI = L\left[\frac{I^2}{2}\right]_0^{I_0} = \frac{1}{2}LI_0^2$$ $$\boxed{U = \tfrac{1}{2}LI_0^2}$$

This energy is stored in the magnetic field of the inductor. It is returned to the circuit when the current falls to zero.

a) Lenz's law

Statement: The induced current always flows in a direction that opposes the change in magnetic flux producing it.

Explanation: When the N pole of a magnet is pushed towards a coil, the flux through the coil increases. The induced current makes the near face of the coil an N pole, which repels the magnet. When the magnet is pulled away, the near face becomes an S pole, which attracts it back. In both cases the induced current opposes the motion. We must therefore do mechanical work to move the magnet, and this work appears as electrical energy. So Lenz's law follows from the law of conservation of energy.

  N[====]S  --->   (( coil ))   near face becomes N : repulsion
  N[====]S  <---   (( coil ))   near face becomes S : attraction

b) Magnet falling through a metal ring

No. Its acceleration is less than $g$.

• As the magnet falls towards the ring, the flux through the ring increases, so a current is induced in the ring. By Lenz's law this current repels the approaching magnet, giving it an upward force.
• After the magnet passes through, the flux decreases. The induced current now attracts the receding magnet, which is again an upward force.

So the magnet feels an upward magnetic force in addition to its weight, and its net downward acceleration is less than $g$. (If the ring is cut, no current flows and the acceleration becomes $g$.)

c) Meaning of the negative sign

The negative sign in $\varepsilon = -N\dfrac{d\phi}{dt}$ is Lenz's law. It shows that the induced emf opposes the change in flux that produces it.

(c) Square of number of turns

For a coil, $L = \dfrac{\mu_0 N^2 A}{l}$, so $L \propto N^2$. It does not depend on the current or on the resistance of the coil.

(a)(i) Lenz's law

The direction of the induced current (or emf) is such that it opposes the change in magnetic flux that produces it.

(a)(ii) Consistency with conservation of energy

If a magnet's north pole is pushed towards a coil, the induced current makes the near face a north pole, which repels the magnet. Work must be done against this repulsion, and that mechanical work appears as electrical energy (heat in the coil). If the induced current had the opposite direction, the magnet would be attracted and accelerate on its own while also producing electrical energy, creating energy from nothing. Lenz's law, as it opposes the cause, avoids this and so agrees with conservation of energy.

(b) Energy stored in an inductor

When the current is $i$, the induced emf is $-L\dfrac{di}{dt}$. The work done by the source against it in time $dt$ is

$$dW = L\,\frac{di}{dt}\,i\,dt = L\,i\,di$$

Total work in raising the current from $0$ to $I$:

$$W = \int_0^I L\,i\,di = \tfrac12 LI^2$$

This is stored as magnetic energy: $U = \tfrac12 LI^2$.

(c) Transformer

Given $V_p = 2200$ V, $V_s = 220$ V, $N_p = 5000$, efficiency $= 90\%$, output power $= 8$ kW.

(i) Turns in the secondary

$$\frac{N_s}{N_p} = \frac{V_s}{V_p} \;\Rightarrow\; N_s = 5000\times\frac{220}{2200} = 500$$

(ii) Input power

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \;\Rightarrow\; P_{\text{in}} = \frac{8}{0.90} = 8.89\ \text{kW}$$

(a) Lenz's law: The direction of an induced emf (or current) is such that it opposes the cause that produces it.

Explanation: If the north pole of a magnet is pushed towards a coil, the induced current flows in the sense that makes the face of the coil towards the magnet a north pole, which repels the magnet. If the magnet is pulled away, the near face becomes a south pole, which attracts it. In both cases the induced current opposes the motion, so work must be done against it. This work appears as electrical energy, so Lenz's law is an expression of the conservation of energy.

(b) Copper has a very low resistivity, so the coil has little resistance. The heat loss $I^2R$ in the coil is small, and the inductance coil is efficient.

(c) The induced emf in the moving rod is

$\varepsilon = Blv = 0.15\times0.5\times2 = 0.15\ \text{V}$

$I = \frac{\varepsilon}{R} = \frac{0.15}{3} = 0.05\ \text{A}$

At constant speed the applied force equals the magnetic force on the rod:

$F = BIl = 0.15\times0.05\times0.5 = 3.75\times10^{-3}\ \text{N}$

(B) to reduce eddy current

The core is made of thin sheets that are insulated from each other. This gives a high resistance to the paths of the eddy currents, so the eddy currents and the heat loss they cause become very small.

(a) Lenz's law and conservation of energy

Statement: The direction of the induced current is always such that it opposes the change in magnetic flux that produces it.

Justification:

• When the N pole of a magnet approaches a coil, the induced current makes the near face of the coil an N pole, which repels the magnet. When the magnet is withdrawn, the near face becomes an S pole, which attracts it back.
• So the magnet is always opposed, and an external agent must do mechanical work to move it.
• This mechanical work is converted into the electrical energy of the induced current, which finally appears as heat in the coil.
• Suppose instead the induced current helped the motion. Then a small push would make the magnet move faster and faster, producing more and more current. We would get unlimited electrical energy without doing any work, which breaks the law of conservation of energy.

Hence Lenz's law is a consequence of the principle of conservation of energy.

(b) Direction of induced current in $C_1$ and $C_2$

   C1                                     C2
  (( ))      N[=========]S   --->        ((((()))))
    |                                        |
   (A)                                      (A)

The magnet moves to the right, away from $C_1$ and towards $C_2$.

• Coil $C_1$: The N pole is moving away, so the flux through $C_1$ decreases. To oppose this, the face of $C_1$ nearest the magnet becomes an S pole and attracts the receding N pole. A face acts as an S pole when the current in it is clockwise as seen from that side. So, viewed from the magnet's side, the current in $C_1$ is clockwise.
• Coil $C_2$: The S pole is approaching, so the flux through $C_2$ increases. To oppose this, the face of $C_2$ nearest the magnet becomes an S pole and repels the approaching S pole. Viewed from the magnet's side, the current in $C_2$ is also clockwise.

Both ammeters deflect only while the magnet is moving. Both induced currents oppose the motion of the magnet: $C_1$ pulls it back and $C_2$ pushes it back.

(i) Meaning of the symbols

• $\varepsilon$ is the induced emf in the coil, in volts.
• $\dfrac{d\varphi}{dt}$ is the rate of change of magnetic flux through each turn of the coil, in Wb s$^{-1}$. $N$ is the number of turns, so $N\dfrac{d\varphi}{dt}$ is the rate of change of flux linkage.

(ii) Significance of the negative sign

The negative sign expresses Lenz's law: the induced emf, and the current it drives, always opposes the change in flux that produces it. This is required by conservation of energy.

(iii) Why ball P takes longer than ball Q

   P o           o Q
     |           |
  +--|------+    |    h
  |x x x x x|    |
  |x x B x x|    |
  +---------+    |
 ==================== ground

• Ball P is a conductor (copper). As it enters and leaves the magnetic field region, the magnetic flux through it changes. This induces eddy currents inside the ball.
• By Lenz's law, these eddy currents oppose the motion that causes them. The field exerts a force on them that acts upward, against the fall. Part of the ball's kinetic energy is also turned into heat by the eddy currents.
• So ball P is slowed down (its acceleration is less than $g$ while it crosses the field region). Ball Q falls freely with acceleration $g$ throughout.

Hence ball P takes longer than ball Q to reach the ground.