An electric motor has 72 turns of wire wrapped on a rectangular coil, of dimensions 3cm by 4cm. Assume that the motor uses 14A of current and that a uniform 0.5T magnetic field exists within the motor.
1)What is the maximum torque delivered by the motor? Answer in units of Nm.
2)If the motor rotates at 3000 rev/min, what is the peak power produced by the motor? Answer in units of W.
any help is appreciated! thanks
Max torque will be at 90 degrees to the plane of B.
Torque= current*turns*area*B
area is in square meters.
To find the maximum torque delivered by the motor, you can use the formula:
Torque = current * turns * area * B
1) First, convert the dimensions of the coil from centimeters to meters:
Length = 3 cm = 0.03 m
Width = 4 cm = 0.04 m
2) Next, calculate the area of the coil:
Area = Length * Width = 0.03 m * 0.04 m = 0.0012 m²
3) Now substitute the values into the formula:
Torque = 14 A * 72 turns * 0.0012 m² * 0.5 T
4) Multiply the values:
Torque = 0.1 Nm
Therefore, the maximum torque delivered by the motor is 0.1 Nm.
To find the peak power produced by the motor, you can use the formula:
Power = Torque * Angular velocity
where the angular velocity is given in radians per second. First, convert the rotational speed from revolutions per minute (rpm) to radians per second:
Angular velocity = (3000 rev/min) * (2π rad/rev) / (60 s/min) = 314.16 rad/s
Now substitute the values into the formula:
Power = 0.1 Nm * 314.16 rad/s
Multiply the values to get the peak power produced:
Power = 31.42 W
Therefore, the peak power produced by the motor is 31.42 W.