Lenz's Law and the Right-Hand Rule

• Lenz's law obeys Newton's third law of motion and the conservation of energy.

• Lenz's law is based on Faraday's law of induction.

Note carefully what is opposed. It is not the flux, and it is not the magnet. It is the change. If the flux is increasing, the induced current acts to reduce it; if the flux is decreasing, the induced current acts to maintain it. The same coil therefore behaves in opposite ways depending only on which direction the flux is moving — which is exactly what the two cases of Sections 11-3 and 11-4 demonstrate.

When the north pole of the magnet is approaching towards the coil, the flux \(\phi\) linking to the coil increases.

1. According to Faraday's law of electromagnetic induction, when \(\phi\) changes, an EMF — and hence a current \(I\) — is induced in the coil, which will create its own magnetic field.
2. According to Lenz's law, this magnetic field created will oppose the increase in \(\phi\) through the coil. This is possible only if the approaching coil side attains north polarity, since similar poles repel each other.
3. Once the magnetic polarity of the coil side is known, the direction of the induced current can easily be determined by applying the right-hand rule.
4. In this case, the current flows in the anticlockwise direction when viewed from the magnet's side.

When the north pole of the magnet is moving away from the coil, the flux \(\phi\) linking to the coil decreases.

1. According to Faraday's law of electromagnetic induction, an EMF — and hence a current \(I\) — is induced in the coil, which will create its own magnetic field.
2. According to Lenz's law, the magnetic field created will oppose the decrease in \(\phi\) through the coil. This is possible only if the approaching coil side attains south polarity, since dissimilar poles attract each other.
3. Once we know the magnetic polarity of the coil side, we can easily determine the direction of the induced current by applying the right-hand rule.
4. In this case, the current flows in the clockwise direction.

For finding the directions of magnetic field or current, the following rule is used.

Applied to a coil. The same rule works in reverse for a solenoid or coil: curl the fingers of the right hand in the direction the current circulates, and the thumb points towards the face that behaves as a north pole. This is the form used in steps 3 and 4 of Sections 11-3 and 11-4.

Two equivalent statements of the same geometry are also in common use:

The thumb rule relates a current to the field it produces. A different problem arises when a conductor sits in an externally applied field: which way does the induced EMF point, and which way does the force act? Fleming's two rules answer these, and the difficulty is remembering which hand goes with which.

Why two rules at all? Because the two situations are physically opposite. In a generator, motion is the cause and current the effect. In a motor, current is the cause and motion the effect. Lenz's law demands that the effect oppose the cause in each case, and swapping hands is precisely what encodes that reversal.

Lenz's law is often presented as an experimental rule about directions. It is more than that: it is a necessary consequence of energy conservation, and the argument is short enough to be worth following in full.

The quantitative statement. Consider a conductor of length \(l\) sliding at velocity \(v\) along frictionless rails in a field \(B\), with the circuit closed through a resistance \(R\). Four steps complete the accounting:

1. The EMF. From Chapter 10, \(e = Blv\).
2. The current. \(i = e/R = Blv/R\).
3. The force. That current sits in the field, so it experiences \(F = Bil = B^{2}l^{2}v/R\) — and by Lenz's law it opposes the motion.
4. The power. To keep the rod moving at constant speed an external agent must supply \(P_{\text{mech}} = Fv\).

The most consequential application of Lenz's law in this book is the back EMF of a motor, introduced in Chapter 1 and now explicable.

When a motor's armature rotates, its conductors move through the field, so by Faraday's law an EMF is induced in them. By Lenz's law that EMF must oppose the cause of the rotation — which is the applied voltage. Hence:

Chapter 1 warned that back EMF is not a loss but the very mechanism of conversion, since \(EI_a\) is the power converted to mechanical form. Lenz's law explains why: an EMF that aided the supply would deliver energy to the shaft and increase the current drawn — energy from nowhere again.

• Lenz's law: the polarity of the induced EMF is such that it produces a current whose magnetic field opposes the change which produces it. It obeys Newton's third law and conservation of energy.

• The negative sign in \(\varepsilon = -N\,\partial\Phi/\partial t\) indicates that the induced EMF and the change in flux have opposite signs.

• What is opposed is the change, not the flux. An approaching magnet is repelled; a receding one is attracted — whichever pole leads.

• Case 1 (N pole approaching): flux increases, coil face becomes N, current anticlockwise.

• Case 2 (N pole receding): flux decreases, coil face becomes S, current clockwise.

• Right-hand thumb rule: thumb along the current, curled fingers give the field. For a coil, curl the fingers with the current and the thumb points to the north face.

• Fleming's right hand → geneRator; Fleming's left hand → moTor. In both, Forefinger = Field, seCond = Current, thuMb = Motion or force.

• For a sliding rod, \(F = B^{2}l^{2}v/R\) opposes the motion and \(P_{\text{mech}} = P_{\text{elec}}\) exactly.

• Back EMF is Lenz's law in a motor: it limits the current to \((V-E)/R_a\) and its product with armature current is the converted power.

• Thinking Lenz's law opposes the flux. It opposes the change in flux. A decreasing flux is helped by the induced current, not opposed.

• Using the wrong hand. geneRator → Right; moTor → lefT. Match the letters and the error disappears.

• Confusing the right-hand thumb rule with Fleming's right-hand rule. The first relates a current to the field it creates; the second relates motion in an external field to the induced current. Different problems entirely.

• Forgetting to state the viewing direction. "Clockwise" is meaningless without saying from which side the coil is viewed. Always specify.

• Treating the opposing force as a loss. The work done against it is the electrical output. Without it there would be no energy conversion at all.

• Believing back EMF wastes power. \(EI_a\) is the converted power. The waste is \(I_a^2R_a\), which back EMF actually reduces by limiting the current.

• Expecting eddy-current brakes to hold a stationary load. \(F \propto v\), so the force vanishes at standstill. A friction brake is still needed to park.

• Assuming the induced current always flows. The EMF appears regardless, but current — and hence the opposing force — requires a closed circuit. An open-circuited generator is easy to turn.

• Dropping the minus sign as "just a convention". It is the difference between a working machine and a perpetual motion machine.

• Mixing two direction rules within one problem. Pick one convention and carry it through; most sign errors come from switching halfway.

For direction problems, follow the three-step routine of Section 11-2 and always state the viewing direction.

1. P11.1 The south pole of a magnet is moved away from a coil. Find the polarity of the near coil face, the current direction seen from the magnet, and the force on the magnet.

   Show answer

   Step 1: receding magnet → flux decreases.
   Step 2: to oppose a decrease the coil must attract the departing pole, so the near face must become the opposite pole — a north pole.
   Step 3: a face presenting north to the observer carries current anticlockwise as seen from that side.
   Force: north attracts south, so the magnet is attracted — held back as it leaves. Work must still be done to remove it.

2. P11.2 A rod 0.6 m long slides at 4 m/s on frictionless rails in a field of 1.1 T. Circuit resistance is 0.8 \(\Omega\). Find \(e\), \(i\), \(F\), and verify the power balance.

   Show answer

   \[e = Blv = (1.1)(0.6)(4) = 2.64~\mathrm{V}, \qquad i = \frac{2.64}{0.8} = 3.30~\mathrm{A}\]

   \[F = Bil = (1.1)(3.30)(0.6) = 2.178~\mathrm{N}\]

   \[P_{\text{mech}} = Fv = (2.178)(4) = 8.712~\mathrm{W}, \qquad P_{\text{elec}} = i^{2}R = (10.89)(0.8) = 8.712~\mathrm{W} \;\checkmark\]

3. P11.3 A rod of mass 0.080 kg and length 0.5 m falls on vertical rails in a field of 0.9 T, with circuit resistance 0.4 \(\Omega\). Find the terminal velocity.

   Show answer

   \[v_t = \frac{mgR}{B^{2}l^{2}} = \frac{(0.080)(9.81)(0.4)}{(0.81)(0.25)} = \frac{0.3139}{0.2025} = 1.550~\mathrm{m/s}\]

   At that speed the rod dissipates \(mgv_t = (0.080)(9.81)(1.550) = 1.216\) W as heat in the resistance.

4. P11.4 Which of Fleming's rules applies to (a) an alternator stator conductor, (b) a DC motor armature conductor, (c) a conductor being pushed through a field to generate current? State the hand in each case.

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   (a) Alternator — generating — right hand.
   (b) DC motor — producing force — left hand.
   (c) Generating — right hand.
   The test is always: which quantity is the cause? Motion causing current → right; current causing motion → left.

5. P11.5 A 400 V DC motor has an armature resistance of 0.8 \(\Omega\) and draws 30 A. Find the back EMF, the standstill current, and the converted power.

   Show answer

   \[E = 400 - (30)(0.8) = 400 - 24 = 376~\mathrm{V}\]

   \[I_{\text{standstill}} = \frac{400}{0.8} = 500~\mathrm{A} \quad (16.7\times \text{ the running current})\]

   \[P_{\text{conv}} = EI_a = (376)(30) = 11\,280~\mathrm{W}\]

   Check: input \(= VI_a = (400)(30) = 12\,000\) W and copper loss \(= I_a^{2}R_a = (900)(0.8) = 720\) W, so the converted power is \(12\,000 - 720 = 11\,280\) W \(\checkmark\)

6. P11.6 For the rod of P11.2, find the braking force at 2 m/s and at 8 m/s. What is the force at standstill?

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   Since \(F = B^{2}l^{2}v/R\) is proportional to \(v\), and \(F = 2.178\) N at 4 m/s:

   \[F(2) = 1.089~\mathrm{N}, \qquad F(8) = 4.356~\mathrm{N}, \qquad F(0) = 0\]

   The force vanishes at standstill — no motion, no EMF, no current, no force. This is the defining limitation of every eddy-current brake.

7. P11.7 A vertical wire carries current upward. Using the right-hand thumb rule, state the direction of the magnetic field to the north of the wire and to the south of it.

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   With the right thumb pointing upward (the current), the fingers curl anticlockwise when viewed from above. Therefore, viewed from above, the field points:

   • North of the wire: towards the west.

   • South of the wire: towards the east.

   The field circles the wire, so its direction reverses on opposite sides — which is exactly why the forces on the two sides of a coil act in opposite senses and produce a torque.

8. P11.8 Explain, using Lenz's law, why an open-circuited generator is much easier to turn than one supplying a load.

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   With the circuit open, an EMF is induced but no current flows. With no current there is no \(F = Bil\) force, so nothing opposes the rotation except friction and windage.

   Close the circuit and current flows; by Lenz's law it produces a torque opposing the rotation, and the prime mover must now supply \(P = \omega T\) to overcome it. That mechanical power is exactly the electrical power delivered to the load. The effort required to turn a generator is the load, transmitted magnetically across the air gap.

9. P11.9 A strong magnet is dropped down a long vertical copper tube. Describe and explain its motion. Would an aluminium tube behave differently? What about a plastic one?

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   The magnet falls slowly at a nearly constant speed. Its motion induces circulating currents in the tube wall; by Lenz's law these oppose the change, retarding the magnet. It settles at the terminal velocity where the retarding force equals its weight.

   Aluminium: similar but faster, since aluminium's resistivity is about 1.6 times copper's, so the induced currents — and hence the retarding force — are correspondingly smaller.

   Plastic: no effect at all. Plastic is an insulator, no currents can circulate, and the magnet falls freely under gravity. The effect requires a conductor, not a magnetic material — copper and aluminium are both non-magnetic.

10. P11.10 Suppose the induced current aided rather than opposed the change producing it. Trace the consequences for a magnet pushed towards a coil.

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    The coil would attract the approaching magnet, accelerating it. Faster motion → greater \(\mathrm{d}\Phi/\mathrm{d}t\) → larger EMF → larger current → stronger attraction → faster still. The magnet would accelerate without limit while the coil dissipated ever-increasing \(i^{2}R\) heat, with no source supplying either the kinetic or the electrical energy.

    This is a perpetual motion machine of the first kind, which violates conservation of energy. Lenz's law is therefore not an independent experimental fact but a necessary consequence of energy conservation.