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 fleld $B=2.5T$. When a current is passed through the coil, lt completes $50$ revolutions in one second. the power output of the motor is $1.5kW$. The current rating of the coil should be

• A. 1 A
• B. 2 A
• C. 3 A
• D. 4 A

Answer 1

The correct option is B 2 A

Let $I$ be the current through the coil of 100 turns. Let the magnetic moment $\overrightarrow{M}$ make an angle $\theta$ at an instant of time.
Then, torque on the coil
$|\overrightarrow{\tau }|=MB\sin \theta$
Work done by the torque in each half revolution $=\frac{W}{2}$, where $W$ is the work done in one full revolution.
$\therefore \frac{W}{2}={\int }_0^{\pi }MBsin\theta d\theta =IANB{\int }_0^{\pi }\sin \theta d\theta$
$=2IANB$
$\therefore$work output per revolution $=W=4INAB$
Hence work output per second$=Power=4INABf$
$\therefore 1.5\times {10}^3=4INABf$
$I=\frac{1.5\times {10}^3}{4\times 100\times 150\times {10}^{-4}\times 2.5\times 50}$
$\therefore I=\frac{15\times {10}^2}{15\times 50}=2A$