Exercises
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Exercise : Induced Current Direction (Lenz's Law)
The magnet SN appearing in the diagram constitutes with the coil a system (S).
The magnet can move parallel to the axis.
1. Identify the inductor in the system (S).
The inductor is the magnet (SN), as it is the source of the magnetic field that induces current in the coil.
2. What do we call the other part of the system?
The other part is called the induced circuit (the coil in this case).
3. Represent on the diagram the inductor field vector.
The magnetic field vector B points from the North pole to the South pole of the magnet.
4. Is the inductor magnetic field variable for a constant magnet-coil distance? Same question for the magnetic flux?
For a constant distance:
- The magnetic field B is constant (does not vary with time).
- The magnetic flux ΦB is also constant (since both B and the area A remain unchanged).
5. Is there an induction phenomenon as long as the magnet-coil distance remains constant?
No, because induction requires a changing magnetic flux (dΦB/dt ≠ 0). With constant distance, both B and Φ remain constant.
6. Now, we move the south pole of the magnet away from face (1) of the coil.
How does this face behave during the magnet's movement: South or North?
According to Lenz's Law, face (1) will behave as a North pole to attract the removal of the magnet's south pole.
7. On this face, write the appropriate letter (S or N) to indicate the direction of the induced current in the coil.
8. Alternative method to determine induced current direction:
8.1) What happens to the inductor field when moving the magnet away from the coil: increases or decreases?
The magnetic field B through the coil decreases as the magnet moves away.
8.2) According to Lenz's Law, the induced current must create a field in a specific direction. Indicate the direction of Binduced.
Binduced must oppose the decrease, so it points in the same direction as the
original Binductor.
8.3) Knowing the direction of Binduced, show the direction of the induced current on the diagram.
Using the right-hand rule:
- Thumb points in direction of Binduced (toward S of magnet)
- Fingers curl in direction of induced current
Exercise : Electromagnetic Induction
A rectangular coil is placed perpendicular to the magnetic field lines of uniform induction as shown in figure (a). The flux ϕ(t) varies over time according to the graph in figure (b).
The coil has a resistance of 10 Ω and consists of 100 rectangular turns, each 10 cm long and 5 cm wide.
The ends A and B of the coil are connected to a resistor with R = 70 Ω.
1. Verify the appearance of an induced current in specific time intervals
An induced current appears when the magnetic flux changes with time
(dϕ/dt ≠ 0). From the graph, this occurs during:
- 0-10 s (rising flux)
- 10-30 s (constant flux → no current)
- 30-40 s (decreasing flux)
- After 40 s (constant flux → no current)
2. Identify the inductor and the induced element
- Inductor: The source of the magnetic field B (external magnetic field)
- Induced element: The coil where current is induced
3. During [10s;30s] interval
3.1) Calculate the magnetic field B
At t = 10ms, ϕ = 1 Wb (from graph)
ϕ = N·B·A ⇒ B = ϕ/(N·A)
A = length × width = 0.1m × 0.05m = 0.005 m²
B = (1Wb)/(100 × 0.005 m²) =0. 5T
3.2) Deduce the induced emf
During [10ms;30ms], dϕ/dt = 0 ⇒ emf = -N(dϕ/dt) = 0 V
4. Determine induced current direction using Lenz's Law
- 0-10 s: Increasing ϕ ⇒ current opposes increase (clockwise when viewed from side A). i negative.
- 30-40 s: Decreasing ϕ ⇒ current opposes decrease (counter-clockwise from side A). i positive.
5. Faraday's Law expression for induced emf
emf = -N(dϕ/dt)
For 0-10 ms: emf = -100 × (0.8Wb/10 s) = -8 V
For 30-40 ms: emf = -100 × (- 1Wb/10 s) = +10 V
6. Compare Faraday and Lenz results
The signs from Faraday's Law match Lenz's Law predictions:
- Negative emf (0-10 s) ⇒ clockwise current
- Positive emf (30-40 s) ⇒ counter-clockwise current
The two laws are consistent.
7. Determine uAB voltage and plot its time variation
Total resistance = Rcoil + R = 10Ω + 70Ω = 80Ω
Induced current i = emf/Rtotal
- 0-10 ms: i = -8V/80Ω = -0.100 A ⇒ uAB = i×R = -7 V
- 30-40 ms: i = +10V/80Ω = +0.125 A ⇒ uAB = i×R = +8.75 V
Exercise : Electromagnetic Induction
A conducting rod ED moves frictionlessly at velocity V on two parallel horizontal conducting rails spaced ℓ=10cm apart. ED is perpendicular to the rails. A resistor R = 100Ω is connected between points A and C of the rails. The resistance of the rails and rod is negligible. The rod, rails, and resistor form a closed circuit placed in a uniform vertical magnetic field B = 0.2T (Fig.1).
Figure 2 shows the variation of the circuit's surface area S with time.
1. Determine S = f(t) for the three phases (a), (b), and (c)
From the graph:
- Phase (a) 0-20ms: Linear increase ⇒ S(t) = k₁t
- Phase (b) 20-50ms: Constant ⇒ S(t) = S₀
- Phase (c) 50-70ms: Linear decrease ⇒ S(t) = S₀ - k₂(t-50ms)
2. For each phase determine:
a) Magnetic flux expression
Φ = B·S(t)·cosθ = B·S(t) (since θ=0°)
- Phase (a): Φ(t) = B·k₁t
- Phase (b): Φ = B·S₀ (constant)
- Phase (c): Φ(t) = B(S₀ - k₂(t-50ms))
b) Induced emf value
e = -dΦ/dt
- Phase (a): e = -B·k₁ (constant negative)
- Phase (b): e = 0 (no flux change)
- Phase (c): e = B·k₂ (constant positive)
c) Current intensity and direction
I = e/R
- Phase (a): I = -B·k₁/R (clockwise)
- Phase (b): I = 0
- Phase (c): I = B·k₂/R (counter-clockwise)
3. Expression for emf e(B, ℓ, V)
e = B·ℓ·V (motional emf)
Where V is the rod's velocity perpendicular to B and ℓ
4. Velocity vector value and direction in each phase
From e = B·ℓ·V ⇒ V = e/(B·ℓ)
- Phase (a): V = -k₁/(B·ℓ) (positive x-direction)
- Phase (b): V = 0
- Phase (c): V = k₂/(B·ℓ) (negative x-direction)
5. Current direction in phase (b) using Lenz's Law
In phase (b), the flux is constant (dΦ/dt=0), so no current is induced (I=0).
6. Oscilloscope display when R is replaced
With oscilloscope connected directly:
- Voltage will show the induced emf directly
- Phase (a): Constant negative voltage
- Phase (b): Zero voltage
- Phase (c): Constant positive voltage
Scale: 0.05V/div vertical, 20ms/div horizontal
The duration of each segment matches the time intervals (20ms, 30ms, 20ms).
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Exercise: Determining Coil Characteristics (L, r)
A coil with length l = 50 cm has N = 2000 turns, each with surface area S = 10 cm². The coil has inductance L and resistance r.
Part A: Determining Resistance (r)
Method I: Experimental Setup
The coil is connected in series with:
- A resistor R = 48 Ω
- A milliammeter (current measurement)
When the south pole of a magnet approaches face C of the coil, the milliammeter shows a steady current i = 1 mA.
1) Current Direction Using Lenz's Law
Determine the direction of induced current through resistor R when:
- South pole approaches face C
- Current reading is 1 mA
2) Time-Varying Magnetic Field Analysis
The magnetic field strength B varies with time as shown:
a) Magnetic Flux Expression
Write the expression for magnetic flux Φ through the coil in terms of N, B, and S:
b) Induced EMF Calculation
Calculate the induced electromotive force (emf) using the B(t) graph:
c) Coil Resistance Determination
Using the measured current and calculated emf, determine the coil's resistance r:
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