Question

A mild steel ring having a cross sectional area of 500mm2
and a mean circumference of 400mm
has a coil of 200 turns wound uniformly around it. Calculate
a. The reluctance of the ring.
b. The current required to produce a flux of 800µWb in the ring.

Answer 1

Answer to Question #82732, Physics / Electromagnetism

A mild steel ring having a cross sectional area of $500 \, \text{mm}^2$ and a mean circumference of $400 \, \text{mm}$ has a coil of 200 turns wound uniformly around it. Calculate

a. The reluctance of the ring.

Solution.

The reluctance:

$$R = \frac {L}{A \mu \mu_ {0}}$$

where

$$L = 400 \, \text{mm} = 0.4 \, \text{m}$$ $$A = 500 \, \text{mm}^2 = 5 \cdot 10^{-4} \, \text{m}^2$$ $$\mu_ {0} = 4\pi \cdot 10^{-7} \, \text{N/A}^2$$ $$\mu = 1500 \text{ for mild steel}$$

Answer:

$$R = \frac {0.4}{5 \cdot 10^{-4} \cdot 4\pi \cdot 10^{-7} \cdot 1500} = 42.44 \, \text{H}^{-1}$$

b. The current required to produce a flux of $800 \, \mu\text{Wb}$ in the ring.

Solution.

The current:

$$I = \frac {\Phi}{A \mu_ {0} n}$$

where

$$\Phi = 800 \, \mu\text{Wb} = 8 \cdot 10^{-4} \, \text{Wb}$$

Answer to Question #82732, Physics / Electromagnetism

$n = 200$ turns

Answer:

$$I = \frac {8 \cdot 10^{-4}}{5 \cdot 10^{-4} \cdot 4\pi \cdot 10^{-7} \cdot 200} = 6.366 \cdot 10^{3} \, \text{A}$$

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