Ask question

Ask question

All categories

3 years ago

6

A solenoid of 2100 turns, area 10 cm2, and length 30 cm carries a current of 4.0 A. (a) Calculate the magnetic energy stored in

the solenoid from 1/2 LI 2. J [2 points] 0 attempt(s) made (maximum allowed for credit

1 answer:

nikitadnepr [17]3 years ago

3 0

Answer:

E = 0.1472 J

Explanation:

Given that,

The number of turns in the solenoid, N = 2100

Area of the solenoid, A = 10 cm² = 0.001 m²

The length of the solenoid, l = 30 cm = 0.3 m

Current in the solenoid, I = 4 A

We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :

$E=\dfrac{1}{2}LI^2$

Where

L is self inductance of the solenoid,

$L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H$

So,

$E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J$

So, 0.1472 J of energy is stored in the solenoid.

You might be interested in

How do defy gravity?

kaheart [24]

Exert force upward.
Like when you pick something up from the floor, or walk up the stairs.

6 0

4 years ago

Different environments can have very different weather. Which environment is most likely to have the highest humidity?

Vsevolod [243]

Answer:

swamp

Explanation:

I think. The desert is hot and less humid. Mountain is more cold. Tundra...no

5 0

3 years ago

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp

Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha

The conservation momentum states that:

pi = pf

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

m1 = m, m2 = 4m, q1 = e, q2 = 2e

and v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0

3 years ago

The speed of a 2 kg mass on a spring is 5 m/s as it passes through its equilibrium position. What is its frequency if the amplit

Lena [83]

Solution :

Given :

Mass of the object attached to the spring, m = 2 kg

Velocity of the object as it moves, V = 5 m/s

Amplitude of the object as it swings, A = 2.5 m

We have to find the frequency of the object.

We known,

$V_{max} = \omega A = 2 \pi f A$

Therefore, $f= \frac{V_{max}}{2 \pi A}$

$f= \frac{5}{2 \times 3.14 \times 2.5}$

f = 0.32 Hz

Therefore the frequency of the object is 0.32 hertz when the amplitude of the 2 kg mass is 2.5 m moving a speed of 5 m/s.

3 0

3 years ago

Early in 1981 the Francis Bitter National Magnet Laboratory at M.I.T. commenced operation of a 3.3 cm diameter cylindrical magne

Igoryamba

Answer:

The maximum electric field $E_{max}= 0.132V/m$

Explanation:

From the question we are told that

The diameter is $d = 3.3 cm = \frac{3.3}{100} = 0.033m$

The magnetic field of the cylinder is $B = 30 T$

The frequency is $f = 15Hz$

The radial distance is $d_r = 1.4cm = \frac{1.4}{100} = 0.014m$

This magnetic field can be represented mathematically as

$B(t) = B_i + B_1sin (wt + \o_i)$

The initial magnetic field is the average between the variation of the magnetic field which is represented as

$B_i = \frac{30 + 29.6}{2}$

$= 29.8T$

Then $B_1$ is the amplitude of the resultant field is mathematically evaluated as

$B_1 = \frac{30.0 - 29.6}{2}$

$= 0.200T$

The electric field induced can be represented mathematically as

$E = \frac{1}{2} [\frac{dB }{dt} ]d_r$

$= \frac{d_r}{2} \frac{d}{dt} (B_i + B_1 sin (wt + \o_o))$

$= \frac{1}{2} (B wr cos (wt + \o_o))$

At maximum electric field $cos (wt + \o_o) = 1$

$E_{max} = \frac{1}{2} B_1 wd_r$

$E_{max} = \frac{1}{2} B_1 2 \pi f d_r$

$= \frac{1}{2} (0.200) (2 \pi (15 ))(0.014)$

$E_{max}= 0.132V/m$

6 0

3 years ago