Question

The rotor in a certain electric motor is a flat, rectangular coil with 80 turns of wire and dimensions 2.50 cm by 4.00 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 10.0 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 3.60×10³ rev/min. (b) Find the peak power output of the motor.

Answer 1

The Peak power output of the motor is 0.452 Watt

To find the peak power output, the given values are,

No .of turns of the wire = 80

Dimensions is given as, 2.50 cm by 4.00 cm

Magnetic field = 0.800 T.

current = 10 mA.

Angular speed = 3.60 ×10³ rev/min

What is Peak power output?

• Peak power output can be also known as Peak work rate.
• If the output was greatest or the work production was very high in the given amount of time then it is known as Peak power output.
• This peak power is depends on the force, distance and time.

The formula for peak power output,

                     Pout = τ * ω

where,

Pout - output power  watts (W),

τ - torque   Newton meters (Nm),

ω - angular speed  radians per second (rad/s).

Calculating angular speed as here rotational speed of the motor in rpm is given:

                      ω = rpm * 2π / 60

where,

ω – angular speed, radians per second (rad/s),

rpm – rotational speed in revolutions per minute,

π – mathematical constant pi (3.14),

60 – number of seconds in a minute.

So, the formula is,

Peak power output = τ *rpm * 2π / 60

τ = NIAB sinθ Nm

  = 80×10×10⁻³×0.0250×0.040×0.800× Sin 90°

  = 2 × 6.4 × 10 ⁻⁴

  = 1.2 × 10⁻⁴ Nm.

Peak power output = 1.2 × 10⁻⁴ ×3.60×10³ × 2π / 60

= 0.452 Watt

Thus, the peak power output is 0.452 Watt.

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