Question

Suppose a 52-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T

Answers

  1. Answer:

    Magnitude of induced emf = 124.8 V

    The induced Emf is directed in the plane of the loop, in the x direction; perpendicular to the coil’s motion (if we agree the motion of the coil is in the y direction)

    Explanation:

    According to Faraday’s law of electromagnetic induction,

    Induced emf = -N (dΦ/dt)

    N = number of turns = 52

    Φ = magnetic flux = BA

    B = magnetic field strength

    A = Cross sectional Area of the loop

    (dΦ/dt) = (d/dt) BA

    Since, B is constant and A is variable,

    (dΦ/dt) = B (dA/dt)

    Induced emf = – NB (dA/dt)

    Induced emf = – NB (ΔA/t)

    N = 52 turns

    B = 1.60 T

    ΔA = 0 – 0.150 = – 0.15 m²

    t = 0.100 s

    Induced emf = – (52)(1.60)(-0.15/0.1)

    Induced emf = 124.8 V

    For the direction of the induced Emf, Fleming’s left hand rule explains that if the index finger points toward magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

    Hence, the induced Emf is directed in the plane of the loop, in the x direction; perpendicular to the coil’s motion

    Hope this helps!!!

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