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1a. States Faradays laws of electro-magnetic induction. 1b. From the second law shown that the induce emf is given by e = N(Φ2 – Φ1) / t. 1c. State the following laws: i. Fleming’s Right Hand Rule ii. Lenz’s law

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1a. States Faraday's laws of electro-magnetic induction.
Faraday's laws of electromagnetic induction describe how a changing magnetic field creates an electromotive force (EMF) in a conductor.

  • First Law: Whenever a conductor is placed in a varying magnetic field, an electromotive force (EMF) is induced. If the conductor circuit is closed, a current is induced.
  • Second Law: The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linkage with the circuit.

1b. From the second law shown that the induce emf is given by e=N(Φ2−Φ1)te = \frac{N(\Phi_2 - \Phi_1)}{t}.
Faraday's second law states that the induced EMF (ee) is proportional to the rate of change of magnetic flux linkage.
Magnetic flux linkage is the product of the number of turns (NN) and the magnetic flux (Φ\Phi) through each turn.
So, flux linkage =NΦ= N\Phi.

  1. Express the change in magnetic flux linkage.

    If the magnetic flux changes from an initial value Φ1\Phi_1 to a final value Φ2\Phi_2 over a time interval tt, the change in magnetic flux linkage is NΦ2−NΦ1=N(Φ2−Φ1)N\Phi_2 - N\Phi_1 = N(\Phi_2 - \Phi_1).

  2. Apply Faraday's second law.

    The induced EMF is proportional to the rate of change of magnetic flux linkage:

    e∝N(Φ2−Φ1)te \propto \frac{N(\Phi_2 - \Phi_1)}{t}

    Introducing a constant of proportionality (which is 1 in SI units, and the negative sign indicates Lenz's law direction, but for magnitude, we often consider the absolute value):

    e=N(Φ2−Φ1)te = \frac{N(\Phi_2 - \Phi_1)}{t}

    This formula represents the average induced EMF over the time interval tt.

    1c. State the following laws:

    • i. Fleming's Right Hand Rule: This rule is used to determine the direction of induced current or EMF when a conductor moves in a magnetic field. If the thumb, forefinger, and middle finger of the right hand are held mutually perpendicular, and the forefinger points in the direction of the magnetic field, the thumb points in the direction of motion of the conductor, then the middle finger points in the direction of the induced current.
    • ii. Lenz's Law: This law states that the direction of the induced current or EMF is always such that it opposes the change in magnetic flux that produced it. It is a consequence of the principle of conservation of energy.

    2a. A coil of 500 turns in link with a flux of 4 mWb. If this flux is reversed in 8 ms, calculate the average emf induced in the coil.

  3. Identify the given values and convert units.

    Number of turns, N=500N = 500
    Initial magnetic flux, Φ1=4 mWb=4×10−3 Wb\Phi_1 = 4 \text{ mWb} = 4 \times 10^{-3} \text{ Wb}
    Since the flux is reversed, the final magnetic flux, Φ2=−4 mWb=−4×10−3 Wb\Phi_2 = -4 \text{ mWb} = -4 \times 10^{-3} \text{ Wb}
    Time taken for reversal, Δt=8 ms=8×10−3 s\Delta t = 8 \text{ ms} = 8 \times 10^{-3} \text{ s}

  4. Calculate the change in magnetic flux.

    ΔΦ=Φ2−Φ1=(−4×10−3 Wb)−(4×10−3 Wb)=−8×10−3 Wb\Delta \Phi = \Phi_2 - \Phi_1 = (-4 \times 10^{-3} \text{ Wb}) - (4 \times 10^{-3} \text{ Wb}) = -8 \times 10^{-3} \text{ Wb}

  5. Calculate the average induced EMF using Faraday's law.

    The magnitude of the average induced EMF is given by:

    e=N∣ΔΦ∣Δte = N \frac{|\Delta \Phi|}{\Delta t}

    Substitute the values:

    e=500×∣−8×10−3 Wb∣8×10−3 se = 500 \times \frac{|-8 \times 10^{-3} \text{ Wb}|}{8 \times 10^{-3} \text{ s}} e=500×8×10−38×10−3e = 500 \times \frac{8 \times 10^{-3}}{8 \times 10^{-3}} e=500×1e = 500 \times 1 e=500 Ve = \boxed{\text{500 V}}

    2b. A coil of 500 turns is wound on a magnetic circuit of reluctance 200000 AT/Wb. If a current of 2A flowing in the coil is reversed in 10 ms, find the average emf induce in the coil.

  6. Identify the given values and convert units.

    Number of turns, N=500N = 500
    Reluctance, S=200000 AT/WbS = 200000 \text{ AT/Wb}
    Initial current, I1=2 AI_1 = 2 \text{ A}
    Since the current is reversed, the final current, I2=−2 AI_2 = -2 \text{ A}
    Time taken for reversal, Δt=10 ms=10×10−3 s\Delta t = 10 \text{ ms} = 10 \times 10^{-3} \text{ s}

  7. Calculate the initial and final magnetic flux.

    The magnetic flux (Φ\Phi) in a magnetic circuit is given by Φ=MMFS=NIS\Phi = \frac{MMF}{S} = \frac{NI}{S}.
    Initial flux:

    Φ1=NI1S=500×2 A200000 AT/Wb=1000200000 Wb=0.005 Wb\Phi_1 = \frac{N I_1}{S} = \frac{500 \times 2 \text{ A}}{200000 \text{ AT/Wb}} = \frac{1000}{200000} \text{ Wb} = 0.005 \text{ Wb}

    Final flux:

    Φ2=NI2S=500×(−2 A)200000 AT/Wb=−1000200000 Wb=−0.005 Wb\Phi_2 = \frac{N I_2}{S} = \frac{500 \times (-2 \text{ A})}{200000 \text{ AT/Wb}} = \frac{-1000}{200000} \text{ Wb} = -0.005 \text{ Wb}

  8. Calculate the change in magnetic flux.

    ΔΦ=Φ2−Φ1=(−0.005 Wb)−(0.005 Wb)=−0.01 Wb\Delta \Phi = \Phi_2 - \Phi_1 = (-0.005 \text{ Wb}) - (0.005 \text{ Wb}) = -0.01 \text{ Wb}

  9. Calculate the average induced EMF using Faraday's law.

    The magnitude of the average induced EMF is given by:

    e=N∣ΔΦ∣Δte = N \frac{|\Delta \Phi|}{\Delta t}

    Substitute the values:

    e=500×∣−0.01 Wb∣10×10−3 se = 500 \times \frac{|-0.01 \text{ Wb}|}{10 \times 10^{-3} \text{ s}} e=500×0.010.01e = 500 \times \frac{0.01}{0.01} e=500×1e = 500 \times 1 e=500 Ve = \boxed{\text{500 V}}

    2c. An air cored solenoid has 600 turns its length is 50 cm and its cross section is 3 cm². calculate the self- inductance in henry.

  10. Identify the given values and convert units to SI.

    Number of turns, N=600N = 600
    Length of the solenoid, l=50 cm=0.50 ml = 50 \text{ cm} = 0.50 \text{ m}
    Cross-sectional area, A=3 cm2=3×(10−2 m)2=3×10−4 m2A = 3 \text{ cm}^2 = 3 \times (10^{-2} \text{ m})^2 = 3 \times 10^{-4} \text{ m}^2
    Permeability of free space (for air core), μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \text{ H/m}

  11. Use the formula for self-inductance of a solenoid.

    The self-inductance (LL) of an air-cored solenoid is given by:

    L=μ0N2AlL = \frac{\mu_0 N^2 A}{l}

  12. Substitute the values and calculate.

    L=(4π×10−7 H/m)×(600)2×(3×10−4 m2)0.50 mL = \frac{(4\pi \times 10^{-7} \text{ H/m}) \times (600)^2 \times (3 \times 10^{-4} \text{ m}^2)}{0.50 \text{ m}} L=4π×10−7×360000×3×10−40.50 HL = \frac{4\pi \times 10^{-7} \times 360000 \times 3 \times 10^{-4}}{0.50} \text{ H} L=4π×3.6×105×3×10−4×10−70.50 HL = \frac{4\pi \times 3.6 \times 10^5 \times 3 \times 10^{-4} \times 10^{-7}}{0.50} \text{ H} L=4π×10.8×10−60.50 HL = \frac{4\pi \times 10.8 \times 10^{-6}}{0.50} \text{ H} L=135.7168×10−60.50 HL = \frac{135.7168 \times 10^{-6}}{0.50} \text{ H} L=271.4336×10−6 HL = 271.4336 \times 10^{-6} \text{ H} L≈2.71×10−4 HL \approx 2.71 \times 10^{-4} \text{ H} L=0.271 mHL = \boxed{\text{0.271 mH}}
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