All categories

3 years ago

What is an advantage of using an electromagnet rather than a regular magnet

Physics

2 answers:

Lena [83]3 years ago

5 0

Electromagnet would have a stronger pull than a regular magnet
<span />

SVEN [57.7K]3 years ago

4 0

Answer:

Electromagnets are temporary magnets made by supplying electric current to a soft iron wounded with coils of wire. Once the source of electricity is disconnected, the magnetism dies off.

However electromagnets usually have stronger electric fields than regular magnets, therefore they are often used in lifting large piece of equipment from one point to another, provided that the equipment are magnetic in nature.

Another major advantage of using an electromagnet rather than a regular magnet is that the magnetism of the electromagnet can be terminated when need be by simply disconnecting the source of electricity.

For example in lifting a heavy metal from one point to another if it gets to the point of dropping off the metal there will be a need to terminate the magnetism so that the metal can fall off; a regular magnet can not be used for this purpose.

You might be interested in

Which galaxy type is smooth and uniform in appearance, where no new stars are forming?

Oksi-84 [34.3K]

Answer: Elliptical galaxy

Explanation:

There are mainly three types of galaxies: Spiral, Elliptical and Irregular. Elliptical galaxies do not have any arms unlike spiral galaxies. The elliptical galaxies are smoother and uniform in appearance. These contain relatively old stars with any site of new star formation missing. In fact, scientists consider elliptical galaxies as those galaxies which are nearing the end of their evolution.

6 0

3 years ago

French jrnf;kwnb;vnkj2wert

dangina [55]

Answer:

???

Explanation:

4 0

3 years ago

Which statement is true? CTP hydrolysis can be a source of chemical energy. dATP is the most common source of chemical energy in

Gemiola [76]

1. Statement A is true.

When GTP is hydrolysed, the free energy of hydrolyses is used to power or drive reactions that are favourable energetically.

2. Statement B it true.

ATP is a complex chemical that gives energy for the activities in many living cells. During hydrolyses, chemical energy stored in the energy-rich phosphoanhydride is released. Hence its a common source of chemical energy in cells.

3. Statement C is false.

The hydrolyses of ATP to ADP in the presence of phosphate, releases one mole of ATP which is estimated to be -57Kj/mol not 14Kj/mol. Below is the equation;

ATP + H20 -----> ADP + Pi + Free energy.

4. Statement D is false.

GTP stands for guanosine triphosphate.

ATP stands for adenosine triphosphate

ADP stands for adenosine diphosphate.

3 0

3 years ago

What is the circle of least confusion?

Artist 52 [7]

Answer and Explanation:

In optics, a CoC(Circle of Confusion) is defined the minimum cross section of a paraxial bundle of rays made by a lens which is sphero-cylindrical type and can be viewed as an optical spot, which do not converge perfectly at the focus while a point source is being imaged due to spherical aberration.

The Circle of Confusion is also referred to as circle of indistinctness or a blur spot

5 0

3 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago