An is a heterogeneous mixture in which visible particles settle

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

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 $=3.7\ MA$

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.

8 0

3 years ago

Drew wants to convert square mils to circular mils in an area calculation. How does he do this?

bija089 [108]

Answer:

i dont know

Explanation:

blah blah blah your welcome

6 0

3 years ago

Helppppppp pleaseeee

SVETLANKA909090 [29]

Answer: (Sorry, but I don't know how to calculate mass)

1. 15 N

2. 0.4921 $\frac{ft}{s^2}$ (feet per second squared)

4. 150 N

5. 8.202 feet per second squared

3 0

2 years ago

Which describes what an outside observer at rest would observe about a spaceship that speeds up and approaches the speed of ligh

Art [367]

The outside observer, at rest relative to the spaceship, would see the spaceship
get shorter. and the clocks on the spaceship run slower than they should.

At the same time, the crew of the spaceship, looking back at the observer on
Earth, would see the observer on Earth get shorter, and the observer's clock
run slower than it should.

They would both be measuring what they see correctly.

6 0

3 years ago

Read 2 more answers

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by ' $m_{best}$ '