Question

(a) Using the values given for an MHD drive in Exercise \({\rm{22}}{\rm{.59}}\), and assuming the force is uniformly applied to the fluid, calculate the pressure created in \({\rm{N/}}{{\rm{m}}^{\rm{2}}}\). (b) Is this a significant fraction of an atmosphere?

Short answer

1. The pressure obtained is:\({\rm{1}}{\rm{.0 x 1}}{{\rm{0}}^{\rm{3}}}{\rm{ N/m}}\).
2. No, it is not a significant fraction of atmosphere.

Step-by-step solution

Define Magnetism

Property of a substance, due to which it undergoes magnetic interactions with other magnetic substances and moving charges, is known as magnetism.

Property of a substance, due to which it undergoes magnetic interactions with other magnetic substances and moving charges, is known as magnetism.

(a)

To evaluate the pressure\(\left( P \right)\), we use the equation:\(P = \frac{F}{A}\)

The value\(F\)is the force and\({\rm{A}}\)is the area which is equal to\(\pi {r^2}\).

\({\rm{P = }}\frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^{\rm{2}}}}}\)

\({\rm{ = }}\frac{{{\rm{50}}\;{\rm{N}}}}{{{\rm{\pi \times (0}}{\rm{.125}}\;{\rm{m}}{{\rm{)}}^{\rm{2}}}}}\)

\({\rm{ = 1}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{3}}}\,{\rm{N/m}}\)

Therefore, the pressure is: \({\rm{1}}{\rm{.0 x 1}}{{\rm{0}}^{\rm{3}}}{\rm{ N/m}}\).

Explaining is it significant fraction of an atmosphere

(b)

Comparison of the pressure we got with the atmospheric pressure is:

\(\frac{{\rm{P}}}{{{{\rm{P}}_{{\rm{atm}}}}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ N/m}}}}{{{\rm{1}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{ N/m}}}}\)

\({\rm{ = 0}}{\rm{.01}}\)

Therefore, pressure is not a significant fraction of the atmospheric pressure.