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schepotkina [342]
3 years ago
15

What is required for the maximum high tide to occur?.

Physics
1 answer:
daser333 [38]3 years ago
4 0

Answer:

Tides are very long waves that move across the oceans. They are caused by the gravitational forces exerted on the earth by the moon, and to a lesser extent, the sun. ... Because the gravitational pull of the moon is weaker on the far side of the Earth, inertia wins, the ocean bulges out and high tide occurs.

Explanation:

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to
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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

b) Current at the center of the circular coil =2.48\times 10^{5}\ A

c) Current near the center of a solenoid 236.8\ A

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,

Conversion 1\times 10^6 A = 1\ MA,and 2cm=\frac{2}{100}=0.02\ m

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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