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koban [17]
4 years ago
8

Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?

Physics
1 answer:
Aneli [31]4 years ago
6 0

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

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Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

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I = (Bπ) (0.17195R⁴)

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