Ask question

Ask question

All categories

3 years ago

9

A 8.0-cm long solenoid has 2000 turns of wire and carries a current of 5.0-A. Calculate the strength of the magnetic field at th

e center of the solenoid in teslas.

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

B = 0.157 T

Explanation:

Given that,

Length of the solenoid, l = 8 cm = 0.08 m

Number of turns in the wire, N = 2000

Current, I = 5 A

We need to find the strength of the magnetic field at the center of the solenoid. It is given by the formula as follows :

$N=\mu_o nI$ , N is number of turns per unit length of solenoid.

So,

$B=4\pi \times 10^{-7}\times \dfrac{2000}{0.08}\times 5\\\\=0.157\ T$

So, the magnetic field at the center of the solenoid is 0.157 T.

You might be interested in

A charge Q accumulates on the hollow metallic dome, of radius R, of a Van de Graaff generator. A probe measures the electric fie

frez [133]

Answer:

E'=(3/16)E

Explanation:

The electric field generated by a Van de Graff generator can be taken as the electric field generated by a hollow shell (for distance out of the shell):

$E=k\frac{Q}{R^2}$

Q: electric charge

K: Coulomb's constant

R: distance in which E is measured.

If the distance is increased to R'=2R, and the charge to Q'=(4/3)Q the new electric field is:

$E'=k\frac{Q'}{R'^2}=k\frac{(3/4)Q}{(2R)^2}=\frac{3}{4}k\frac{Q}{4R^2}=\frac{3}{16}k\frac{Q}{R^2}=\frac{3}{16}E$

hence, the new electric field E' is 3/16 times the previous electric field E

7 0

3 years ago

Read 2 more answers

Arrange the following types of electromagnetic radiation in order of increasing (lowest to highest) frequency from left to right

algol13

Electromagnetic radiation in order of increasing (lowest to highest) frequency is Radio waves < Green visible light < UV radiation.

<h3>What is Frequency?</h3>

This is defined as the rate at which something occurs or is repeated over a particular period of time. The unit of frequency is referred to as Hertz.

Electromagnetic radiations with increasing frequency can be seen below:

Radio waves, microwaves, infrared waves, visible light, ultraviolet radiations, X rays, γ rays.

Read more about Frequency here brainly.com/question/254161

4 0

2 years ago

suppose 384g of steam originally at 100C is quickly cooled to produce liquid water at 31C. How much heat must be removed from th

dlinn [17]

Answer:

$Q=977216.256\ J=977.216\ kJ$

Explanation:

Given:

mass of steam, $m=384\ g$

temperature of steam, $T_{is}=100^{\circ}C$

temperature of resultant water, $T_{fw}=31^{\circ}C$

We have,

latent heat of vapourization of water, $L=2256\ J.g^{-1}$

specific heat capacity of water, $c=4.186\ J.g^{-1}$

<em>When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.</em>

<u>Now the heat removed from steam to achieve the final state of water:</u>

$\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water$

$Q=m(L+c.\Delta T)$

$Q=384(2256+4.186\times (100-31))$

$Q=977216.256\ J=977.216\ kJ$

3 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b

fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² = 376 J

After coming to a flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0

3 years ago