Question

An electric motor is a device to convert electrical energy into mechanical energy. The motor shown below has a rectangular coil (\[15cm \times 10cm\]) with 100 turns placed in a uniform magnetic field \[B = 2.5T\]. When a current is passed through the coil. It completes 50 revolutions in one second. The power output of the motor is \[1.5kW\]. The current rating should be

A. 1 A
B. 2 A
C. 3 A
D. 4 A

Answer 1

Hint: We need to calculate the work done by the coil due to the current in the wire. The rate of change of this work done is the power output of the motor.

Formula used: In this solution we will be using the following formulae;
\[W = - 4mB\] where \[W\] is the work done in a cycle of the electric motor, \[m\] is the magnetic moment due to the current, \[B\] is the magnetic field.
\[m = - NIA\] where \[N\] is the number of turns of the coil, \[I\] is the current flowing through the coil, and \[A\] is the area of the loop of the coil.

Complete step by step answer:
To calculate the current, we need to first find the expression of work done with respect to the current.
Generally, the work done in one cycle of an electric motor can be given as
\[W = - 4mB\] where \[W\] is the work done in a cycle of the electric motor, \[m\] is the magnetic moment due to the current, \[B\] is the magnetic field.
But the magnetic moment can be given as
\[m = - NIA\] where \[I\] is the current flowing through the coil, and \[A\] is the area of the loop of the coil.
Hence, we have
\[W = 4NIAB\]
In 50 revolutions, it would be
\[W = 4NIAB \times 50\]
The power which is work done in one second, would be
\[\dfrac{W}{{1s}} = P = 4NIAB \times \dfrac{{50}}{{1s}}\]
So, we make current subject, we have
\[I = \dfrac{P}{{4NAB\left( {50} \right)}}\]
Hence, by inserting all values, we get
\[I = \dfrac{{1500}}{{4\left( {100} \right)\left( {0.15 \times 0.1} \right)\left( {2.5} \right)\left( {50} \right)}} = 2A\]
Hence, the correct answer is B.

Note: For clarity, generally, the power of the electric motor can just be written directly as \[P = 4NIAB\omega \] where \[\omega \] is the frequency of revolution (in revolution per second).