All categories

3 years ago

Does a thicker core make an electromagnet stronger?

Engineering

2 answers:

mel-nik [20]3 years ago

6 0

Answer:

Yes

Explanation:

The core of an electromagnet serves to stabilize the magnetic field created by the wire. The thicker the core, the more metal there is to amplify the current. Therefore, a thicker core does make an electromagnet stronger. Hope this helps!

iogann1982 [59]3 years ago

3 0

My conclusion is that the electromagnet with the thick core is stronger than the electromagnet with the thin core. Wire size did not make a significant difference in electromagnet strength.

You might be interested in

Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase

ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ( $\% V_{Gr}$ ) is determined by the following expression:

$\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%$

Where:

$V_{Gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{Fe}$ - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

$V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}$

$V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}$

Where:

$m_{Gr}$ , $m_{Fe}$ - Masses of the graphite and ferrite phases, measured in grams.

$\rho_{Gr}$ , $\rho_{Fe}$ - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

$\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%$

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

$m_{Gr} = \frac{2.5}{100}\times (100\,g)$

$m_{Gr} = 2.5\,g$

$m_{Fe} = 100\,g - 2.5\,g$

$m_{Fe} = 97.5\,g$

If $m_{Gr} = 2.5\,g$ , $m_{Fe} = 97.5\,g$ , $\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}$ and $\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}$ , the volume percent of graphite is:

$\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%$

$\% V_{Gr} = 91.906\,\%$

The volume percent of graphite is 91.906 per cent.

6 0

3 years ago

It is illegal to improperly dispose of antifreeze.<br> True<br> False

Reil [10]

Ehheem✔️

Explanation:
✔️✖️✔️✖️✔️✖️

6 0

3 years ago

Read 2 more answers

What do you think of web 3.0? do you think it will be realized someday in the future?

Elodia [21]

Answer:

is this a question for hoework

7 0

3 years ago

1.0•10^-10 standard form

Drupady [299]

Answer:

$1.0 * 10^{-10} = 0.0000000001$

Explanation:

Given

$1.0 * 10^{-10}$

Required

Convert to standard form

$1.0 * 10^{-10}$

From laws of indices

$a^{-x} = \frac{1}{a^x}$

So, $1.0 * 10^{-10}$ is equivalent to

$1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}$

$1.0 * 10^{-10} = 1.0 * 0.0000000001$

$1.0 * 10^{-10} = 0.0000000001$

Hence, the standard form of $1.0 * 10^{-10}$ is $0.0000000001$

3 0

3 years ago

Make two lists of applications of matrices, one for those that require jagged matrices and one for those that require rectangula

Agata [3.3K]

Answer:

Explanation:

You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.

3 0

3 years ago