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nadezda [96]
3 years ago
10

A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive dir

ection. What is the magnitude of the minimum magnetic field needed to levitate the rod
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

Magnetic field, B = 0.23 T          

Explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.

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I attached the full question.
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Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
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6 0
3 years ago
How does the amount of friction affect the sum of forces
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Answer:

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Explanation:

<h3>mark as brainliast</h3>

6 0
2 years ago
Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the cur
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Answer:

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Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

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8 0
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