The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The rotor rotates

Question

The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The rotor rotates in a uniform magnetic field of 0.800 T. When the plane of the rotor is perpendicular to the direction of the magnetic field, the rotor carries a current of 10.5 mA. In this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn steadily at an angular speed of 3.54 103 rev/min.(a) Find the maximum torque acting on the rotor. ______ N · m

(b) Find the peak power output of the motor.______W

(c) Determine the amount of work performed by the magnetic field on the rotor in every full revolution.______ J

(d) What is the average power of the motor?______ W

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Thiên Hương 4 years 2021-08-22T17:59:44+00:00 1 Answers 199 views 0

Answers ( )

  1. Nho
    0
    2021-08-22T18:01:24+00:00
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    Given Information:

    Number of turns = N = 84

    Area of Rectangular coil = 2.61×3.64 cm = 0.0261×0.0364 m

    Magnetic field = B = 0.80 T

    Current = I = 10.5 mA = 0.0105 A

    Angular speed = ω = 3.54×10³ rev/min

    Required Information:

    (a) Maximum torque = τmax = ?

    (b) Peak output power = Ppeak = ?

    (c) Work done = W = ?

    (d) Average power = Pavg?

    Answer:

    (a) Maximum torque = 0.00067 N.m

    (b) Peak output power = 0.248 W

    (c) Work done = 0.00189 J

    (d) Average power = 0.1115 W

    Explanation:

    (a) The toque τ acting on the rotor is given by,

    τ = NIABsin(θ)

    Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

    A = 0.0261×0.0364

    A = 0.00095 m²

    The maximum toque τ is achieved when θ = 90°

    τmax = NIABsin(90°)

    τmax = 84*0.0105*0.00095*0.80*1

    τmax = 0.00067 N.m

    (b) The peak output power of the motor is given by,

    Pmax = τmax*ω

    ω = 3.54×10³ x 2π/60

    ω = 370.7 rad/sec

    Pmax = 0.00067*370.7

    Pmax = 0.248 W

    (c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

    W = 2∫NIABωsin(ωt) dt

    W = -2NIABcos(ωt)

    Evaluating limits,

    W = -2NIABcos(π) – (-2NIABcos(0))

    W = 2NIAB + 2NIAB

    W = 4NIAB

    W = 4*84*0.0105*0.00067*0.80

    W = 0.00189 J

    (d) Average power of the motor is given by

    Pavg = W/t

    t = 2π/ω

    t = 2π/370.7

    t = 0.01694 sec

    Pavg = W/t

    Pavg = 0.00189/0.01694

    Pavg = 0.1115 W

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