22.2.4 Induced emf in a Linear Conductor
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| Figure 22.30 A linear conductor is connected to a galvanometer and placed between two permanent magnets with opposite poles. When the conductor moves and cuts the magnetic field lines, an induced emf is generated, causing the galvanometer pointer to deflect. |
1. Inducing an emf in a linear conductor: a demonstration
Connect a linear conductor to a galvanometer. Place the conductor horizontally in between a pair of permanent magnets with opposite poles facing each other, as shown in Figure 22.30. Perform the following steps and observe the deflection, if any, of the pointer of the galvanometer.
| Procedure | Observation | Deduction |
|---|---|---|
| Do not move the conductor in the magnetic field. | The pointer does not deflect at all. | No emf is induced in the conductor. |
| Move the conductor horizontally so that it does not ‘cut’ the field. | The pointer does not deflect. | No emf is induced if the conductor does not ‘cut’ the field. |
| Move the conductor vertically upwards quickly so that it ‘cuts’ the field. | The pointer of the galvanometer is observed to deflect to one side. | An induced current flows in the conductor when the conductor ‘cuts’ the field. |
| Move the conductor vertically downwards quickly so that it ‘cuts’ the field. | The pointer is observed to deflect to the opposite side. | An induced current flows through the conductor, but now it flows in the opposite direction. |
| Move the conductor vertically upwards or downwards at different speed. | The pointer deflects through larger angle for higher speed. | The induced current is larger for higher speed. |
| Repeat using conductor of different lengths moving vertically at the same speed. | The pointer deflects through larger angle for longer length. | The induced current is larger for longer conductor. |
| Repeat using different magnetic field strength, each time moving the conductor vertically at the same speed. | The pointer deflects through larger angle for stronger field. | The induced current is larger for stronger field. |
2. Equation for motional emf
Using the apparatus shown above, we discover that when the conductor moves and ‘cuts’ a magnetic field, an emf is induced in it. This emf induced as a result of motion in a magnetic field is referred to as the motional emf. Results obtained from experiments indicate that the motional emf E is dependent upon the following quantities:
E ∝ B (magnetic flux density of the magnetic field)
E ∝ L (length of conductor perpendicular to field)
E ∝ v⊥ (velocity of conductor perpendicular to field)
From these experimental results we obtain the following relationship:
E ∝ BL(v⊥)
or
E = k BL(v⊥)
If we let B = 1 T, L = 1 m and v⊥ = 1 m s-1, and define the motional emf induced under those stated conditions as 1 V, then we get k = 1.
E = BL(v⊥)
Suppose that the conductor does not ‘cut’ the field perpendicularly.
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| Figure 22.31 A straight conductor moves with velocity v in a magnetic field B, making an angle θ with the field. Only the velocity component v⊥ = v sinθ that is perpendicular to the magnetic field contributes to the motional emf induced in the conductor. |
Instead, it moves at velocity v which makes an angle θ with the field, as shown in Figure 22.31. Then the velocity component v⊥ perpendicular to the field is given by
v⊥ = v sin θ
Hence, the motional emf is given by
E = BLv sin θ
3. Induced current and its direction
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| Figure 22.32 The thumb, index finger, and middle finger of the right hand are held mutually perpendicular to each other. The thumb represents the direction of motion (velocity, v), the index finger represents the direction of the magnetic field (B), and the middle finger indicates the direction of the induced current (I) in the conductor. |
We can use a rule, known as Fleming’s right hand rule, to help us determine the direction of the induced current flowing through the linear conductor. The rule:
(a) Point the thumb of the right hand upwards, the index finger forward and the middle finger sideway, mutually perpendicular to the thumb and the index finger, as shown in Figure 22.32.
(b) The thumb points in the direction of the velocity (v) and the index finger points in the direction of the magnet field (B).
(c) Then the middle finger points in the direction of the induced current (I).
EXAMPLE 22.18
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| Figure 22.33 A uniform metal rod PQ is connected to an external resistor and placed perpendicular to a uniform magnetic field directed out of the page. When the rod moves at a constant velocity, it cuts the magnetic field lines, inducing an emf and causing a current to flow in the circuit. |
A uniform metal rod PQ of length 1.0 m and diameter 1.0 mm is connected to an ohmic resistor which has 2.0 Ω of resistance. It is then placed inside and perpendicular to a uniform magnetic field, as shown in Figure 22.33. The field has strength 0.50 T and points out of the page. The rod is made to move at a constant speed of 1.5 m s-1 in the direction which is perpendicular to the field and indicated by the arrow shown.
(a) Determine
(i) the resistance of the rod
(ii) the induced current flowing in the circuit
(iii) the electrical power dissipated by the circuit.
(b) Determine the direction of current flow in the circuit.
(Resistivity of the metal rod = 5.0 × 10-7 Ω m)
Answer
(a) (i) Cross-sectional area of rod
A = 1/4 πd2
= 1/4 π (1.0 × 10-3)2 = 7.86 × 10-7 m2
Resistance of rod
R = ρL / A
= (5.0 × 10-7)(1.0) / 7.86 × 10-7
= 0.64 Ω
(ii) Induced emf
Induced current
E = BLv sin 90°
= (0.50)(1.0)(1.5) = 0.75 V
I = E
R + R1
= 0.75
0.64 + 2.0 = 0.28 A
(iii) Electrical power dissipated
P = IE
= (0.28)(0.75) = 0.21 W
(b) Using Fleming’s right hand rule, we find that the induced current flows in the metal rod from Q to P.
EXAMPLE 22.19
An airplane flies towards the east at a speed of 200 m s−1. The Earth’s magnetic field, which is of strength 25 μT, points at an angle of 60° to the horizontal.
(a) Determine the induced emf developed across the tips of the wing, separated by a distance of 15 m.
(b) Which end of the tip of the wing is at a potential higher than that at the other tip?
Answer
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| Figure 22.34 Front view of an airplane flying east in the Earth’s magnetic field. The vertical component of the magnetic field induces an emf between the tips of the wings. |
(a) Figure 22.34 shows the front view of the airplane. The vertical component of the strength of the Earth’s magnetic field is given by
Bv = B sin 60° = 25 sin 60° = 21.7 μT
Induced emf
E = BvLv sin 90°
= (21.7 × 10−6)(15)(200) = 6.5 × 10−2 V
(b) (i) Suppose that the tips X and Y are connected to a wire to form a closed circuit. Then an induced current would flow in the wire as well as through the airplane, from one tip to the other.
(ii) Use the Fleming’s right hand rule to determine the direction of the induced current along the wings of the airplane. The induced current flows from X to Y along the airplane, and flows in the wire from Y to X, as shown in Figure 22.35.
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Figure 22.35 Direction of induced current along the airplane wings determined using Fleming’s right-hand rule. The current flows from X to Y along the wing. |
(iii) Since the current flows out of Y, it means that tip Y is positively charged while tip X is negatively charged.
EXAMPLE 22.20
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| Figure 22.36 A metal rod XY slides on a U-shaped conductor PQRS placed in a uniform magnetic field B directed into the page. |
A metal rod XY of resistance R can slide without friction on a U-shaped conductor PQRS, as shown in Figure 22.36. The resistance of the U-shaped conductor may be neglected. The conductor and rod are placed in a uniform magnetic field of strength B which points into the page. PQRS is placed on a horizontal surface.
(a) An agent pushes XY at constant speed v towards QR. Show that an emf E induced in XY is given by
E = BLv
where L is the length of rod XY.
(b) Is the end X positively or negatively charged relative to end Y? Explain.
(c) Derive an expression for the force which must be exerted on the rod by the agent in order to move the rod at constant speed in terms of E, B, R and L.
(d) Derive an expression for the electrical power dissipated by the circuit in terms of B, L, v, R.
(e) Show that energy is conserved when the agent moves the rod at constant speed towards QR.
Answer
(a) Initial magnetic flux linkage through the closed loop is
Φi = NΦ
= NBA cos θ
= (1)(B)(Ai) cos 0°
= BAi
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| Figure 22.37 A conducting rod moves along a U-shaped conductor, reducing the area of the closed loop over time. This change in area causes a decrease in magnetic flux linkage, which induces an emf in the circuit according to Faraday’s law. |
After time t the rod has moved through distance x and the area of the loop has become smaller, as shown in Figure 22.37. The final magnetic flux linkage through the closed loop is
Φf = BAf
Change of flux linkage
ΔΦ = Φf − Φi = B(Af − Ai) = B(Lx)
Rate of change of flux linkage
ΔΦ
t
= B(Lx)
t
= BL(x
t)
= BLv
where v is the constant speed of the rod. The induced emf E is given by
E = − ΔΦ
t
= −BLv
(b) The loop is a closed circuit. Hence, a current flows through the rod XY because an induced emf exists across XY. Using Fleming’s right hand rule, we deduce that the current flows in the direction XYRQX. Since current flows into X from the loop, it means that X must be negatively charged relative to Y.
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| Figure 22.38 When current flows through the moving rod inside a magnetic field, a magnetic force acts on the rod. According to Fleming’s left-hand rule, this force opposes the motion, and an external force is required to maintain constant velocity. |
(c) When current flows in the rod, which is inside a magnetic field, a magnetic force F′ acts on the rod. Using Fleming’s left hand rule, we can deduce that this force acts to the right since the induced current flows from X to Y, as shown in Figure 22.38. Since the rod is forced to move at constant speed, the net force acting on the rod must be zero. Hence, the magnitude of the force F exerted by the agent on XY must be
F = F′
But we have
F′ = BIL sin 90°
where I is the induced current. Hence
F = BIL = B(E
R)L
(d) The electrical power Pe dissipated by the circuit is given by
Pe = E2
R
= (BLv)2
R
(e) The mechanical power Pm expended by the agent is given by
Pm = Fv = (BIL)v = B(E
R)Lv
= B(BLv
R)Lv
= (BLv)2
R
We have Pm = Pe
provided that no energy is dissipated due to friction forces. This means that the mechanical energy expended per second by the agent is completely converted to electrical energy. Hence, energy is conserved.
Applications of Induced emf in a Linear Conductor
The concept of induced emf in a linear conductor has many practical applications in daily life and technology. Below are some important applications:
| Application | Description | Example |
|---|---|---|
| Electric Generator | Mechanical energy is converted into electrical energy when a conductor moves in a magnetic field. | Power plants (hydroelectric, wind turbines) |
| Electric Motor (Back emf) | Moving conductors inside motors produce induced emf that opposes the applied voltage. | Fans, washing machines |
| Magnetic Flow Meter | Measures the flow rate of conducting fluids using induced emf. | Water supply systems, medical devices |
| Railgun / Electromagnetic Launch | Uses motion of conductors in magnetic fields to generate large forces. | Military and experimental launch systems |
| Speed Sensors | Induced emf depends on velocity, so it can be used to measure speed. | Bicycle dynamo, tachometer |
| Aircraft Wing emf | Moving aircraft in Earth's magnetic field produces small emf across wings. | Aviation instrumentation |
Conclusion: The magnitude of induced emf depends on magnetic field strength, conductor length, and velocity. This principle is widely used in energy generation, measurement devices, and modern engineering systems.
Summary: Induced emf in a Linear Conductor
An induced electromotive force (emf) is produced when a conductor moves through a magnetic field and cuts magnetic field lines. This phenomenon is known as motional emf.
Key Concepts
- No motion → No emf
- Conductor must cut magnetic field lines
- Direction determined by Fleming’s Right-Hand Rule
Main Formula
E = BLv sin θ
- B = magnetic field strength
- L = length of conductor
- v = velocity of conductor
- θ = angle between motion and magnetic field
Important Observations
- Higher speed → Larger induced emf
- Longer conductor → Larger induced emf
- Stronger magnetic field → Larger induced emf
- Direction of motion affects current direction
Energy Concept
Mechanical energy used to move the conductor is converted into electrical energy. If no energy is lost:
Mechanical Power = Electrical Power
Applications
- Electric generators
- Dynamo / speed sensors
- Aircraft wing emf
- Magnetic flow meters
Conclusion: Induced emf depends on magnetic field strength, conductor length, and motion, and is widely used in modern electrical technology.
Conclusion
The induced electromotive force (emf) in a linear conductor occurs when the conductor moves through a magnetic field and cuts the magnetic field lines. The magnitude of the induced emf depends on three main factors: the strength of the magnetic field, the length of the conductor, and the velocity of motion.
The relationship is given by E = BLv sin θ, which shows how these factors combine to determine the value of the induced emf. The direction of the induced current can be determined using Fleming’s Right-Hand Rule.
This principle demonstrates the conversion of mechanical energy into electrical energy. When energy losses are negligible, the mechanical power supplied is equal to the electrical power produced, showing that energy is conserved.
Overall, induced emf in a linear conductor is a fundamental concept in electromagnetism and plays an important role in many real-world applications such as electric generators, sensors, and modern electrical systems.
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