Answer to Question #142965 in Electrical Engineering for Ahmed

Question

2 mWb is to be produced in the air gap of the magnetic circuit shown in figure 21.17. How much ampere turns the coil must provide to achieve this? Relative permeability μr of the core material may be assumed to be constant and equal to 5000. All the dimensions shown are in cm and the sectional area is 25cm2 through out.

Answer 1

Since the magnetic flux is $\Phi=B\times S$ , where $B$ is magnetic induction, $S$ is the cross-sectional area, $L=\frac{\Phi}{I}=\frac{\mu_0\times \mu \times N \times S}{l}$ , where $I$ is the current, $\mu$ is the relative permeability of the substance, $\mu_0$ is the magnetic constant, $N$ the number of turns, and therefore $N\times I=\frac{\Phi\times l}{\mu \times \mu_0\times S}$ $=\frac{2\times 10^{-3}\times 10{-2}\times(20+20+20-0,1+15+15+15+15)}{5000 \times of 1.26\times 10{-6}\times 25\times 10^{-2}}\approx1,52$