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A rectangular coil of 150 turns, initially 5 cm by 8 cm, changes its dimensions to 7 cm by 11 cm in 0.15 s. If the magnetic field is 25 T and is 30 degrees to the plane of the coil, what is the induce

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2 parts
a

a) E=46.25 V\text{a) } \mathcal{E} = \text{46.25 V}

b

b) E=3.209 J\text{b) } E = \text{3.209 J}

Step-by-step solution

7 steps

Here's the solution for Question 3:

Part a· 5 steps

Induced emf

  1. Identify given values and convert units.

    Number of loops, N=150N = 150
    Initial length, l1=5 cm=0.05 ml_1 = 5 \text{ cm} = 0.05 \text{ m}
    Initial width, w1=8 cm=0.08 mw_1 = 8 \text{ cm} = 0.08 \text{ m}
    Final length, l2=7 cm=0.07 ml_2 = 7 \text{ cm} = 0.07 \text{ m}
    Final width, w2=11 cm=0.11 mw_2 = 11 \text{ cm} = 0.11 \text{ m}
    Time interval, Δt=0.15 s\Delta t = 0.15 \text{ s}
    Magnetic field strength, B=25 TB = 25 \text{ T}
    Angle relative to the plane of the coil =30∘= 30^\circ.

  2. Calculate initial and final areas, and the change in area.

    Initial area:

    A1=l1×w1=0.05 m×0.08 m=0.004 m2A_1 = l_1 \times w_1 = 0.05 \text{ m} \times 0.08 \text{ m} = 0.004 \text{ m}^2

    Final area:

    A2=l2×w2=0.07 m×0.11 m=0.0077 m2A_2 = l_2 \times w_2 = 0.07 \text{ m} \times 0.11 \text{ m} = 0.0077 \text{ m}^2

    Change in area:

    ΔA=A2−A1=0.0077 m2−0.004 m2=0.0037 m2\Delta A = A_2 - A_1 = 0.0077 \text{ m}^2 - 0.004 \text{ m}^2 = 0.0037 \text{ m}^2

  3. Determine the angle for magnetic flux calculation.

    The angle θ\theta in the magnetic flux formula ΦB=BAcos⁡θ\Phi_B = B A \cos \theta is the angle between the magnetic field vector and the normal to the coil's plane. Since the magnetic field is 30∘30^\circ relative to the plane of the coil, the angle with the normal is:

    θ=90∘−30∘=60∘\theta = 90^\circ - 30^\circ = 60^\circ

  4. Calculate the change in magnetic flux.

    The change in magnetic flux is due to the change in area:

    ΔΦB=BΔAcos⁡θ\Delta \Phi_B = B \Delta A \cos \theta ΔΦB=25 T×0.0037 m2×cos⁡(60∘)\Delta \Phi_B = 25 \text{ T} \times 0.0037 \text{ m}^2 \times \cos(60^\circ) ΔΦB=25×0.0037×0.5=0.04625 Wb\Delta \Phi_B = 25 \times 0.0037 \times 0.5 = 0.04625 \text{ Wb}

  5. Calculate the induced emf using Faraday's Law.

    Faraday's Law of Induction states:

    E=−NΔΦBΔt\mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t} E=−150×0.04625 Wb0.15 s\mathcal{E} = -150 \times \frac{0.04625 \text{ Wb}}{0.15 \text{ s}} E=−150×0.308333... V\mathcal{E} = -150 \times 0.308333... \text{ V} E=−46.25 V\mathcal{E} = -46.25 \text{ V}

    The magnitude of the induced emf is:

    a) E=46.25 V\boxed{\text{a) } \mathcal{E} = \text{46.25 V}}

Part b

b) E=3.209 J\text{b) } E = \text{3.209 J}

Energy dissipated in the circuit

  1. Calculate the power dissipated.

    Given the total internal resistance, R=100 ohmsR = 100 \text{ ohms}.
    The power dissipated in the circuit is given by:

    P=E2RP = \frac{\mathcal{E}^2}{R} P=(46.25 V)2100 ohmsP = \frac{(46.25 \text{ V})^2}{100 \text{ ohms}} P=2139.0625 V2100 ohms=21.390625 WP = \frac{2139.0625 \text{ V}^2}{100 \text{ ohms}} = 21.390625 \text{ W}

  2. Calculate the total energy dissipated.

    The energy dissipated over the time interval Δt\Delta t is:

    E=P×ΔtE = P \times \Delta t E=21.390625 W×0.15 sE = 21.390625 \text{ W} \times 0.15 \text{ s} E=3.20859375 JE = 3.20859375 \text{ J}

    Rounding to four significant figures:

    b) E=3.209 J\boxed{\text{b) } E = \text{3.209 J}}
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