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MariettaO [177]

2 years ago

13

Review. Consider the deuterium-tritium fusion reaction with the tritium nucleus at rest:

1 answer:

asambeis [7]2 years ago

7 0

Considering the deuterium-tritium fusion reaction with the tritium nucleus at rest: ¹₂H + ¹₃H → ²₄He + ⁰₁n the electric potential energy (in electron volts) at this distance is 17.58MeV

<h3>How is the electric potential energy of deuterium-tritium fusion reaction calculated?</h3>

The reaction is ¹₂H + 1₃H → ²₄He + ⁰₁n

Value of Q = (Mass of ¹₂H + Mass of ¹₃H - Mass of ²₄He- Mass of n) x 931 MeV

Mass of ¹₂H = 2.014102

Mass of ¹₃H = 3.016049

Mass of ²₄He = 4.002603

Mass of n = 1.00867

Therefore Value of Q = [2.014102+3.016049−4.002603−1.00867] × 931 MeV

Therefore Value of Q = 0.01887 × 931 MeV

= 17.58MeV

To learn more about deuterium-tritium fusion reaction, refer

brainly.com/question/9054784

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jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

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Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

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$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

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