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horsena [70]
3 years ago
9

Students hypothesized that by running an electric current through the wire of the apparatus shown here, they could cause a non-m

agnetic nail to exhibit magnetic properties. What would be a reasonable way to test this?
a

Question 7 options:

a.Weigh the nail first while before turning the current on and again while the current was on.


b.Put the nail in water before turning the current on and again after turning the current off.


c.Examine the nail under a microscope before turning the current on and again after turning the current off.


d.See if iron filings will stick to the nail before turning the current on and again while the current was on.
Physics
2 answers:
astra-53 [7]3 years ago
5 0
The answer is d i think
Ivenika [448]3 years ago
5 0

The answer is D i think because i got it right!

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Which of these is a likely impact of stronger than normal trade winds on the eastern Pacific Ocean?
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Answer:

Jet stream would be displaced southwards causing heavy rain and flooding.

Explanation:

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What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J
Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)

and on the surface the current is

I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A

(b)

For Jb = J0(1 - r/R)

I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )

and on the surface

I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0

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Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi  \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A

and in a simmilar way for Jb

I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi   J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

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Answer:

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