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A square coil with 2000 turns and a side length of 150 mm is placed in a magnetic field. The magnetic flux density changes from 0.4 T to 0.2 T in 0.8 s. If the coil's resistance is 4 ohms and the angl

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2 parts
1

Magnitude of induced EMF=7.95 V\text{Magnitude of induced EMF} = 7.95 \text{ V}

2

Magnitude of induced current=1.99 A\text{Magnitude of induced current} = 1.99 \text{ A}

Step-by-step solution

4 steps

Here are the calculations for the induced EMF and induced current.

Given information:

  • Number of turns, N=2000N = 2000
  • Side length of the square coil, L=150 mm=0.15 mL = 150 \text{ mm} = 0.15 \text{ m}
  • Initial magnetic flux density, B1=0.4 TB_1 = 0.4 \text{ T}
  • Final magnetic flux density, B2=0.2 TB_2 = 0.2 \text{ T}
  • Time interval, Δt=0.8 s\Delta t = 0.8 \text{ s}
  • Total resistance of the coil, R=4ΩR = 4 \Omega
  • Angle relative to the field, θ=45∘\theta = 45^\circ

Part 1· 3 steps

Calculate the magnitude of the induced EMF.

  1. Calculate the area of the square coil.

    A=L2A = L^2 A=(0.15 m)2A = (0.15 \text{ m})^2 A=0.0225 m2A = 0.0225 \text{ m}^2

  2. Calculate the change in magnetic flux density.

    ΔB=B2−B1\Delta B = B_2 - B_1 ΔB=0.2 T−0.4 T\Delta B = 0.2 \text{ T} - 0.4 \text{ T} ΔB=−0.2 T\Delta B = -0.2 \text{ T}

    For magnitude, we use ∣ΔB∣=0.2 T|\Delta B| = 0.2 \text{ T}.

  3. Calculate the magnitude of the induced EMF using Faraday's Law of Induction.

    The formula for induced EMF is E=−NΔΦΔt\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}, where Φ=BAcos⁡θ\Phi = BA \cos \theta.
    Since the magnetic flux density changes, ΔΦ=ΔB⋅A⋅cos⁡θ\Delta \Phi = \Delta B \cdot A \cdot \cos \theta.
    For the magnitude of EMF:

    ∣E∣=N∣ΔB∣⋅A⋅cos⁡θΔt|\mathcal{E}| = N \frac{|\Delta B| \cdot A \cdot \cos \theta}{\Delta t} ∣E∣=2000×0.2 T×0.0225 m2×cos⁡(45∘)0.8 s|\mathcal{E}| = 2000 \times \frac{0.2 \text{ T} \times 0.0225 \text{ m}^2 \times \cos(45^\circ)}{0.8 \text{ s}} ∣E∣=2000×0.2×0.0225×0.707106780.8|\mathcal{E}| = 2000 \times \frac{0.2 \times 0.0225 \times 0.70710678}{0.8} ∣E∣=2000×0.00318198050.8|\mathcal{E}| = 2000 \times \frac{0.0031819805}{0.8} ∣E∣=2000×0.0039774756|\mathcal{E}| = 2000 \times 0.0039774756 ∣E∣=7.9549512 V|\mathcal{E}| = 7.9549512 \text{ V}

    Rounding to three significant figures:

    Magnitude of induced EMF=7.95 V\boxed{\text{Magnitude of induced EMF} = 7.95 \text{ V}}

Part 2

Magnitude of induced current=1.99 A\text{Magnitude of induced current} = 1.99 \text{ A}

Calculate the magnitude of the induced current.

  1. Use Ohm's Law to find the induced current.

    I=∣E∣RI = \frac{|\mathcal{E}|}{R} I=7.9549512 V4ΩI = \frac{7.9549512 \text{ V}}{4 \Omega} I=1.9887378 AI = 1.9887378 \text{ A}

    Rounding to three significant figures:

    Magnitude of induced current=1.99 A\boxed{\text{Magnitude of induced current} = 1.99 \text{ A}}
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