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NASA SP-413

Nov 17, 1975 ... The concept of spacecraft Earth, a sphere of finite resources and ominous ... for ten weeks to construct a convincing picture of how ... Appendix K — The Lunar Gas Gun Mass Driver . ... that space colonization offers large potential benefits and ... kPa or ~100 mm Hg) for good respiration yet low enough to. Qalaxia Master Bot
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Glossary of physics - Wikipedia

This glossary of physics is a list of definitions of terms and concepts relevant to physics, its sub-disciplines, and related fields, including mechanics, materials ... Pravalika
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Given that

Potential of gun collector = 100  V

Magnetic fiels B = 2.83 * 10^{-4} T

Radius of the curved path r = 12.0 cm = 12 *10^2 m

Step 1: Determine the velocity of an ejected electron from gun due to voltage

Kinetic energy of electron K.E = \frac{1}{2} mv^2

v^2 = \frac{2 K.E}{m}

According to energy conversion  K.E = energy of each electron = eV

v^2 = \frac{2 eV}{m}     ....................(1)

Step 2: Velocity of electron in a spherical bulb due to magnetic field.

Electron moving in a circular path so it acquires centripetal force

F = \frac{mv^2}{r}

Force on the electron in a magnetic field

F =evB                            \because \theta = 90\degree

So, from the two equations we can write, evB = \frac{mv^2}{r}

eB = \frac{mv}{r}

v = \frac{eBr}{m}

v^2 = \frac{e^2B^2r^2}{m^2} ....................(2)

Step 3: Calculating the specific charge ratio

From equation (1) and (2)

\frac{e^2B^2r^2}{m} = \frac{2 eV}{m}

\frac{e}{m} = \frac{2V}{B^2 r^2}

Plug in the values in the above equation.

$\frac{e}{m} = \frac{2 * 100}{[2.83 * 10^{-4}]^2 (12*10^2)^2}$

\frac{e}{m} = 1.734 * 10^{11} C/kg

Hence, the specific charge ratio   \frac{e}{m} = 1.734 * 10^{11} C/kg