A ring of mean diameter 30 cm is wound with 200 turns of copper wire carrying a current of 2Â
A. The x-section of the magnetic material of the ring is 12 cm2 and its relative permeability isÂ
1,000. Determine theflux through it.
Diameter(D)=30cm \\
Circumference=\pi\times D \\
C=30\pi cm \\
Turns (N)=200 \\
Current (I)= 2A \\
Relative permeability (\mu_r)=1000 \\
Cross section area (A)=12cm^2 \\
\mu_o= 4\pi \times 10^{-7} N/A^2 \\
\Phi= \frac {\mu_o\times \mu_r A\times N \times I}{C} \\
\Phi= \frac{4\pi \times 10^{-7} \times 1000 \times 12 \times 200 \times 2}{30\pi} \\
\Phi= 32mWb
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