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The maximum working flux density
of a lifting electromagnet is 1.8 T
and the effective area of a pole face
is circular in cross-section. If the
total magnetic flux produced is 353
mWb, then the diameter of the pole
face in meters.
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A

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UpStudy AI · Accepted Answer

1. Given that the magnetic flux is related to the flux density and the effective area by
   $$
   \Phi = B \, A,
   $$
   where $$\Phi$$ is the flux, $$B$$ is the flux density, and $$A$$ is the area.

2. The effective area $$A$$ of the circular pole face is given by
   $$
   A = \pi \left(\frac{d}{2}\right)^2,
   $$
   where $$d$$ is the diameter.

3. Solving for the area using the flux equation, we have
   $$
   A = \frac{\Phi}{B}.
   $$
   The total flux is given as $$353 \, \text{mWb} = 353 \times 10^{-3} \, \text{Wb} = 0.353 \, \text{Wb}$$ and the flux density is $$1.8 \, \text{T}$$. Thus,
   $$
   A = \frac{0.353}{1.8} \approx 0.1961 \, \text{m}^2.
   $$

4. Equate the expressions for $$A$$:
   $$
   \pi \left(\frac{d}{2}\right)^2 = 0.1961.
   $$

5. Solve for $$d$$:
   $$
   \left(\frac{d}{2}\right)^2 = \frac{0.1961}{\pi},
   $$
   $$
   \frac{d}{2} = \sqrt{\frac{0.1961}{\pi}},
   $$
   $$
   d = 2\sqrt{\frac{0.1961}{\pi}}.
   $$

6. Substitute the numerical value for $$\pi \approx 3.1416$$:
   $$
   d = 2\sqrt{\frac{0.1961}{3.1416}} \approx 2\sqrt{0.0624} \approx 2 \times 0.2498 = 0.4996.
   $$

7. Rounding to an appropriate number of significant figures, we get
   $$
   d \approx 0.50 \, \text{m}.
   $$

Thus, the diameter of the pole face is approximately $$0.50\, \text{m}$$.
The diameter of the pole face is approximately 0.50 meters.

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