Question 5.13: A horse-shoe-type iron-core electromagnet is wound with 500 ......
A horse-shoe-type iron-core electromagnet is wound with 500 turns and is required to lift heavy iron bars of 200 kg each time. The area of cross section of each of the poles of the horse-shoe magnet is 0.01 m² . The mean length of the flux path through the electromagnet is 0.5 m. Calculate the value of the exciting current through the coil. The relative permeability of the flux path is 1000.
A weight of 200 kg has to be lifted by both the poles of the electromagnet. Thus, each pole will have to lift a load of 100 kg. The force of attraction which is also called the lifting power of the electromagnet is
\mathrm{F}=\frac{\mathrm{B}^2 \mathrm{~A}}{2 \mu_0}or, \mathrm{B}=\sqrt{\frac{\mathrm{F} \times 2 \mu_0}{\mathrm{~A}}}
Putting the values,
\mathrm{F}=100 \mathrm{~kg}=100 \times 9.8 N
\mathrm{B}=\sqrt{\frac{100 \times 9.8 \times 2 \times 4 \pi \times 10^{-7}}{0.01}}Again = 0.5 Wb/m²
B = μH
or, \mathrm{H}=\frac{\mathrm{B}}{\mu}=\frac{\mathrm{B}}{\mu_0 \mu_{\mathrm{r}}}=\frac{0.5}{4 \pi \times 10^{-7} \times 1000}=400
We know
\mathrm{H} \times l=\mathrm{AT}=\mathrm{NI}or, \mathrm{I}=\frac{\mathrm{H} \times l}{\mathrm{~N}}=\frac{400 \times 0.5}{500}=0.4 \mathrm{~A}