a) The sketch below is of a simple electric motor. A copper disk of radius r = 6

Question

a) The sketch below is of a simple electric motor. A copper disk of radius r = 6.50

a) The sketch below is of a simple electric motor. A copper disk of radius r = 6.50?10-3 m is mounted on a shaft that rests on bearings. Electrical contact is made through sliding brushes at a and b, and current I = 3.50?10-3 A passes through the circuit. A magnetic field B = 9.00?10-2 T is directed into the page. The shaft is connected to a load, on which the motor performs work. If the disk turns at a speed v = 5.60?101 Hz, what is the power output? b) Compute the torque exerted by the motor on its load.

Explanation / Answer

To calculate the force in a magnetic field due to electric conductor we know that
F = BILSin(angle) where F is force , B is magnetic field and I is Current , therefore we will consider an small element of length dL therefore force on that dF = BIdL since angle between length and magnetic field is 90 therefore Sin90= 1.Now we will integrate this over 0 to R where R is radius of disc
F = BI(2(pi)R) , we know the value of every variable hence we can got the value of force.
For calculation of Power we know that
Power = Force*Velocity = Torque*(angular velocity) we know the force and angular velocity is given therefore we can calculate the power.   (Torque = Force*radius of disc)

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