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Convert the velocity from km/hr to m/s.

Physics

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Answer

1.78 mV\text{1.78 mV}

9.6 V\text{9.6 V}

Step-by-step solution

6 steps

1.

  1. Convert the velocity from km/hr to m/s.

    v=80 km/hr×1000 m1 km×1 hr3600 sv = 80 \text{ km/hr} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}} v=800003600 m/s=80036 m/s=2009 m/s≈22.22 m/sv = \frac{80000}{3600} \text{ m/s} = \frac{800}{36} \text{ m/s} = \frac{200}{9} \text{ m/s} \approx 22.22 \text{ m/s}

  2. Convert the magnetic field from microtesla to tesla.

    B=40 µT=40×10−6 TB = 40 \text{ µT} = 40 \times 10^{-6} \text{ T}

  3. Calculate the induced EMF using the formula E=BLv\mathcal{E} = B L v.

    E=(40×10−6 T)×(2 m)×(2009 m/s)\mathcal{E} = (40 \times 10^{-6} \text{ T}) \times (2 \text{ m}) \times \left(\frac{200}{9} \text{ m/s}\right) E=40×10−6×2×2009 V\mathcal{E} = \frac{40 \times 10^{-6} \times 2 \times 200}{9} \text{ V} E=16000×10−69 V\mathcal{E} = \frac{16000 \times 10^{-6}}{9} \text{ V} E=169×10−3 V\mathcal{E} = \frac{16}{9} \times 10^{-3} \text{ V} E≈1.777×10−3 V\mathcal{E} \approx 1.777 \times 10^{-3} \text{ V} E≈1.78 mV\mathcal{E} \approx 1.78 \text{ mV}

    The induced EMF is 1.78 mV\boxed{\text{1.78 mV}}.

    2.

  4. Identify the initial and final magnetic flux.

    Initial magnetic flux Φ1=400 µWb\Phi_1 = 400 \text{ µWb}.
    Since the flux is reversed, the final magnetic flux Φ2=−400 µWb\Phi_2 = -400 \text{ µWb}.

  5. Calculate the change in magnetic flux ΔΦ\Delta \Phi.

    ΔΦ=Φ2−Φ1=(−400 µWb)−(400 µWb)\Delta \Phi = \Phi_2 - \Phi_1 = (-400 \text{ µWb}) - (400 \text{ µWb}) ΔΦ=−800 µWb=−800×10−6 Wb\Delta \Phi = -800 \text{ µWb} = -800 \times 10^{-6} \text{ Wb}

  6. Apply Faraday's Law of Induction, E=−NΔΦΔt\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}.

    Given N=1200N = 1200 turns and Δt=0.1 s\Delta t = 0.1 \text{ s}.

    E=−(1200)×−800×10−6 Wb0.1 s\mathcal{E} = -(1200) \times \frac{-800 \times 10^{-6} \text{ Wb}}{0.1 \text{ s}} E=(1200)×800×10−60.1 V\mathcal{E} = (1200) \times \frac{800 \times 10^{-6}}{0.1} \text{ V} E=1200×8000×10−6 V\mathcal{E} = 1200 \times 8000 \times 10^{-6} \text{ V} E=9600000×10−6 V\mathcal{E} = 9600000 \times 10^{-6} \text{ V} E=9.6 V\mathcal{E} = 9.6 \text{ V}

    The induced EMF is 9.6 V\boxed{\text{9.6 V}}.

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