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erma4kov [3.2K]
2 years ago
11

A solenoid 1.85 m long and 2.20 cm in diameter carries a current of 21.0 A. The magnetic field inside the solenoid is 25.0 mT. F

ind the length of the wire forming the solenoid.
Physics
1 answer:
kobusy [5.1K]2 years ago
8 0

Answer:

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?

Explanation:

hope this helps trying to be brainliest

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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&#10;\\\\&#10;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}&#10;

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)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

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}&#10;

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)&#10;\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}&#10;\\\\B_{out}=\frac{\mu_0ka^3}{3r}&#10;

again using,equaiton 1,

B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}

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3 years ago
Orginize it
tensa zangetsu [6.8K]
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Radiation: Microwave cooking food
- Microwaves use radiation to heat the food inside of it; between radio waves and infrared radiation on the electromagnetic spectrum
Conduction: Touching a hot car seat in the summer
- Conduction is the transfer of heat by direct contact (hand to seat).
Conduction: Burning yourself with a curling iron (Similar to above; direct contact).

Convection: An ocean breeze
- Convection near coastlines cause the transfer of energy; water warms and cools slower than land.
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- You are in direct contact with the slide, which is hot due to the temperature.
Convection: Water in a boiling pot of macaroni
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2 years ago
P2O5 is a covalent compound used to purify sugar. What is the name of this compound?
AfilCa [17]

Answer:

B) Diphosphorus pentoxide

Explanation:

8 0
3 years ago
Read 2 more answers
On February 15, 2013, Asteroid 2012 DA14 passed within 17,200 miles [mi] of the surface of the Earth at a relative speed of 7.8
xz_007 [3.2K]

Answer:

The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ

1 PJ = 10¹⁵ J

Explanation:

Kinetic energy = mv²/2

velocity of the asteroid is given as 7.8 km/s = 7800 m/s

To obtain the mass, we get it from the specific gravity and diameter information given.

Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³

But density = mass/volume

So, mass = density × volume.

Taking the informed assumption that the asteroid is a sphere,

Volume = 4πr³/3

Diameter = 30 m, r = D/2 = 15 m

Volume = 4π(15)³/3 = 14137.2 m³

Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg

Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J

6 0
3 years ago
Convert 141.2 kg to lbs and show all units and work​
zavuch27 [327]
Your answer is 311.29271 lbs
5 0
3 years ago
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