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Afina-wow [57]
3 years ago
11

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

ng the axis and above the center of the loop
Physics
1 answer:
HACTEHA [7]3 years ago
3 0

Answer:

B=2.91\ \mu T

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}

Put all the values,

B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T

So, the required magnetic field is equal to 2.91\ \mu T.

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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____
valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0
3 years ago
What is a physical form in which a substance can exist?
Rama09 [41]
The three physical forms are:

Solid, liquid, or gas.

8 0
3 years ago
According to Newton’s first law of motion, when will an object at rest begin to move?
Natasha_Volkova [10]

Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".

Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed.  So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.

8 0
4 years ago
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Which is a vector quantity speed or velocity
Alecsey [184]
Velocity is a vector quantity. A vector quantity has both a magnitude and a direction. Speed only has a magnitude, but no direction. Velocity has both.
8 0
3 years ago
A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (a) What is the distance from the
Zarrin [17]

The distance from the horizontal top surface of the cube to the water level is "6.282 cm".

<h3>What is Archimedes' principle?</h3>

According to Archimedes' principle, the weight of the fluid that the body displaces is equal to the upward buoyant force that is applied to a body submerged in a fluid, whether fully or partially. The Archimedes' principle is a fundamental physical law in fluid mechanics. It was created by Syracuse's Archimedes.

According to Archimedes' principle, a body submerged in a fluid experiences an upward force proportional to the weight of the fluid that has been displaced. One of the prerequisites for equilibrium is this. We believe that the buoyancy force, also known as the centre of buoyancy, is situated in the middle of the submerged hull.

From Archimedes' principle, we get

\rightarrow L^3 \rho_{\text {Wood }} &=L^2 d \rho_{\text {Water }} \\

d &=L \frac{\rho_{\text {Waat }}}{\rho_{\text {Water }}} \\

&=18 \times \frac{651}{1000} \\

=11.72cm

So,

The distance from horizontal top to the water level will be:

=18-11.72

=6.282cm

To learn more about Archimedes' principle refer to:

brainly.com/question/1155674

#SPJ4

4 0
1 year ago
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