What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N

Answer:

The voltage is equal to the batteries terminal voltage

Explanation:

Explanation:

Answer:

The  rotational kinetic energy is  [tex]K = 2116.3 \ J[/tex]

Explanation:

From the question we are told that

    The moment of inertia  is  [tex]I = 0.35 \ kg \cdot m^2[/tex]

    The number of revolution is N  =  70 revolution

     The  time taken is  t  =  4.0  s

Generally the angular velocity is mathematically represented as

      [tex]w = \frac{2 \pi N }{t }[/tex]

substituting values

      [tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]

       [tex]w = 109.97 \ rad/s[/tex]

The rotational kinetic energy K i mathematically represented as

       [tex]K = \frac{1}{ 2} * I * w^2[/tex]

substituting values

       [tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]

       [tex]K = 2116.3 \ J[/tex]

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

Answer: (V1+V2)/2

Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

Answer:

y = 20.99 V / A

there is no friction    y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight  hydrostatic thrust

              F_net= ∑F = B-W

the expression stop pushing is

              B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

             W = ρ-ice g V

              F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

            F = -dU / dy

            ΔU = - ∫ F dy

            ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

            ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0,  U = 0, that is, the potential energy is zero at the surface

             U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

             Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

             [tex]Em_{f}[/tex] = U = m g y

as there is no friction, energy is conserved

            Em₀ = Em_{f}

            (ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

            ρ_ice = m / V

            m = ρ_ice V

we substitute

             (ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

              y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

             y = 0.913 / (1 - 0.913) 2 V / A

             y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

                V / A = h

where h is the height of the block

                 y = 20.99 h

Answer:

Using a process of testing ideas and gathering evidence

Explanation:

Answer:

counterclockwise

Explanation:

given data

area = 25 cm²

solution

We know that a changing magnetic field induces the current and induced emf is express as

[tex]\epsilon = -N \frac{d \phi }{dt}[/tex]     ..................................1

and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field

so when magnetic field decrease and point coming out of the paper.

so induced current in the loop will be counterclockwise

Answer:

Explanation:

The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;

B is the magnetic field (in Tesla)

I is the current (in amperes)

L is the length of the wire (in meters)

∝ is the angle that the conductor makes with the magnetic field.

Given parameters

L = 0.56 m

I = 2.6A

F = 0.24N

∝  = 90°

Required

magnitude of the magnetic field (B)

Substituting the given values into the formula given above we will have;

F = BILsin∝

0.24 = B * 2.6 * 0.56 sin90°

0.24 =  B * 2.6 * 0.56 (1)

0.24 = 1.456B

1.456B = 0.24

Dividing both sides by 1.456 will give;

1.456B/1.456 = 0.24/1.456

B ≈ 0.165Tesla

Hence the magnitude of the magnetic field is approximately 0.165Tesla

Answer:

A thermos bottle works well because:

A) Its glass walls are thin

Answer:

A thermos bottle works well because:

C

Vacuum reduces heat radiation

Answer:

The speed of the wave is 19.4 m/s

Explanation:

The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m

The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.

this means that the frequency f = 2 Hz

The speed of a wave = fλ = 9.70 x 2 = 19.4 m/s

Answer:

The number of turns of the inductor is 2000 turns.

Explanation:

Given;

emf of the inductor, E = 0.8 V

the rate of change of current with time, dI/dt = 10 A/s

steady current in the solenoid, I = 0.2 A

flux per turn, Ф = 8.0 μWb per

Determine the inductance of the solenoid, L

E = L(dI/dt)

L = E / (dI/dt)

L = 0.8 / (10)

L = 0.08 H

The inductance of the solenoid is given by;

[tex]L = \frac{\mu_o N^2 A}{l}[/tex]

Also, the magnetic field of the solenoid is given by;

[tex]B = \frac{\mu_o NI}{l}[/tex]

I is 0.2 A

[tex]B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}[/tex]

[tex]\frac{B}{0.2 } = \frac{\mu_o N}{l}[/tex]

[tex]L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)[/tex]

But Ф = BA

[tex]L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns[/tex]

Therefore, the number of turns of the inductor is 2000 turns.

This question involves the concepts of magnetic flux, magnetic field, and inductance.

The inductor has "2000" turns.

The magnetic field due to an inductor coil is given as follows:

[tex]B=\frac{\mu_o NI}{L}\\\\[/tex]

where,

B = magnetic field

μ₀ = permeability of free space \

N = No. of turns

I = current = 0.2 A

L = length of inductor

Therefore,

[tex]\frac{\mu_oN}{L}=\frac{B}{0.2\ A}---------- eqn(1)[/tex]

Now, the inductance of a solenoid is given by the following formula:

[tex]E = L\frac{dI}{dt}\\\\L = \frac{E}{\frac{dI}{dt}}[/tex]

The inductance of solenoid can also be given using the following formula:

[tex]L = \frac{\mu_o N^2A}{L}[/tex]

comparing both the formulae, we get:

[tex]\frac{E}{\frac{dI}{dt}}= \frac{\mu_oN^2A}{L}\\\\E=\frac{dI}{dt}\frac{\mu_oN}{l}(NA)\\\\using\ eqn (1):\\\\E=\frac{dI}{dt}\frac{B}{0.2}(NA)\\\\[/tex]

where,

BA = magnetic flux = [tex]\phi[/tex] = 8 μWb/turn = 8 x 10⁻⁶ Wb/turn

N = No. of turns = ?

E = E.M.F = 0.8 volts

[tex]\frac{dI}{dt}[/tex] = rate of change in current = 10 A/s

Therefore,

[tex]0.8=(10)\frac{8\ x\ 10^{-6}}{0.2}N\\\\N=\frac{(0.8)(0.2)}{8\ x\ 10^{-5}}[/tex]

N = 2000 turns

Learn more about magnetic flux here:

brainly.com/question/24615998?referrer=searchResults

The attached picture shows the magnetic flux.

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{1}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{1}=0.01021\ Wb[/tex]

We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{2}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{2}=0.00127\ Wb[/tex]

We need to calculate the magnitude of the average induced emf

Using formula of emf

[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]

Put the value into t5he formula

[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]

[tex]\epsilon=0.016867\ V[/tex]

[tex]\epsilon=16.87\ mV[/tex]

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

[tex]I=\dfrac{\epsilon}{R}[/tex]

Put the value into the formula

[tex]I=\dfrac{0.016867}{2.5}[/tex]

[tex]I=0.00675\ A[/tex]

[tex]I=6.75\ mA[/tex]

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Answer:

The number of interference fringes is  [tex]n = 3[/tex]

Explanation:

From the question we are told that

     The wavelength is  [tex]\lambda = 433 \ nm = 433 *10^{-9} \ m[/tex]

      The distance of separation is  [tex]d = 6 \mu m = 6 *10^{-6} \ m[/tex]

       The  order of maxima is m =  5

The  condition for constructive interference is

       [tex]d sin \theta = n \lambda[/tex]

=>     [tex]\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ][/tex]

=>    [tex]\theta = 21.16^o[/tex]

So at  

      [tex]\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m[/tex]

   [tex]6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}[/tex]

=>    [tex]n = 3[/tex]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

Answer:

8 possible combinations

Assuming R 1, R 2 and R 3 be three different Resistance

1- all three in series

2-all three in parallel

3- R 1 and R 2 in series and parallel with R 3

​

4-R 1 and R 3 in series and parallel with R 2

​

5-R 2 and R 3 in series and parallel with R 1

​

6- R 1 and R 2

in parallel and series with R 3

​

7-R 1 and R 3 in parallel and series with R 2

​

8-R 2 and R 3 in V with R 1

​

Answer:

The wavelength is  [tex]\lambda = 419 \ nm[/tex]

Explanation:

From the question we are told that

   The  distance of separation is   [tex]d = 3.34 *10^{-5} \ m[/tex]

   The  distance of the screen is  [tex]D = 3.30 \ m[/tex]

      The  order of the fringe is  n =  7

     The distance of separation of  fringes is y =  29.0 cm = 0.29 m

Generally the wavelength of the light is mathematically represented as

          [tex]\lambda = \frac{y * d }{ n * D}[/tex]

substituting values

         [tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]

        [tex]\lambda = 4.19*10^{-7}\ m[/tex]

        [tex]\lambda = 419 \ nm[/tex]

Explanation:

Hey, there!

Here, In question given that,

potential difference (V)= 8V

resistance (R)= 10 ohm

Now,

According to the Ohm's law,

V= R×I { where I = current}

or, I = V/R

or, I = 8/10

Therefore, current is 4/5 A or 0.8 A.

(A= ampere = unit of current).

Hope it helps...

Answer:

   v₂ = 9 v

Explanation:

For this exercise in fluid mechanics, let's use the continuity equation

           v₁ A₁ = v₂ A₂

where v is the velocity of the fluid, A the area of ​​the pipe and the subscripts correspond to two places of interest.

The area of ​​a circle is

           A = π R²

let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint

In this case v₁ = v and the area is

            A₁ = π R²

in the second point

           A₂= π (R / 3)²

we substitute in the continuity equation

           v π R² = v₂ π R² / 9

            v = v₂ / 9

            v₂ = 9 v

Answer:

[tex]N = 208 \ turns[/tex]

Explanation:

From the question we are told that

    The  rate of  current change is  [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]

    The  magnitude of the induced emf is  [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]

     The  current is  [tex]I = 1.50 \ A[/tex]

      The  average  flux is  [tex]\phi = 0.00458 \ Wb[/tex]

Generally the number of  turns the number of turn the solenoid has is mathematically represented as  

            [tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]

substituting values

           [tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]

            [tex]N = 208 \ turns[/tex]

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

Complete Question

At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.

Answer

a. Approximately [tex]5*10^{10}[/tex] fissions per second.

b. Approximately [tex]6*10^{12 }[/tex]fissions per second.

c. Approximately [tex]4*10^{11}[/tex] fissions per second.

d. Approximately [tex]3*10^{12}[/tex] fissions per second.

e. Approximately[tex]3*10^{14}[/tex] fissions per second.

Answer:

The correct option is  d

Explanation:

From the question we are told that

       The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]

Thus to generated  [tex]100 \ J/s[/tex] i.e  (100 W  ) the rate of fission is  

              [tex]k = \frac{100}{3.2 *10^{-11} }[/tex]

              [tex]k =3*10^{12} fission\ per \ second[/tex]

Answer:

the correct statement is

* atoms cannot be created or destroyed

Explanation:

The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.

Based on the previous descriptive, the correct statement is

* atoms cannot be created or destroyed

Answer:

the Answer is b hope it help

Explanation:

Answer:

(a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .

Explanation:

Given that,

Width of diffraction grating [tex]w= 19.2\ mm[/tex]

Number of rulings[tex]N=6010[/tex]

Wavelength = 337 nm

We need to calculate the distance between adjacent rulings

Using formula of distance

[tex]d=\dfrac{w}{N}[/tex]

Put the value into the formula

[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]

[tex]d=3.19\times10^{-6}\ m[/tex]

We need to calculate the value of m

Using formula of constructive interference

[tex]d \sin\theta=m\lambda[/tex]

[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]

Here, m = 0,1,2,3,4......

[tex]\lambda[/tex]=wavelength

For largest value of  θ

[tex]\dfrac{m\lambda}{d}>1[/tex]

[tex]m>\dfrac{d}{\lambda}[/tex]

Put the value into the formula

[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]

[tex]m>9.46[/tex]

[tex]m = 9[/tex]

(a). We need to calculate the largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=71.9^{\circ}[/tex]

(b). We need to calculate the second largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=57.7^{\circ}[/tex]

(c). We need to calculate the third largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=47.7^{\circ}[/tex]

Hence, (a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .

Answer:

Not sure but I believe predictable.

Explanation:

Phenomena usually consists of :

- A history, a date in which the physical phenomenon has occurred.

- A source, a place or reason to why or where the physical phenomena has occured.

According to this, I want to say predictable.

It is not repeatable, there are one-time phenomenons that have occurred that scientists to this day still have not recorded again such as the Big Bang.

It is not provable. Most of the theories earlier scientists and historians have predicted have not yet been proved. Yet rather, somehow, they have been explored and investigated.

It is not readable. This is self explanatory, some things scientists investigate are not written down, nor read. It starts with a mental theory and then immediately goes to the next phase of investigation.

Answer:

29.5°

Explanation:

To find the distance d

d = 1E10^-2/8500lines

= 1.17x 10-6m

But wavelength in first order maximum is 577nm

and M = 1

So

dsin theta= m. Wavelength

Theta= sin^-1 (m wavelength/d)

= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6

= 493*10^-3= sin^-1 0.493

Theta = 29.5°

Answer:

The acceleration is   [tex]a =2\ m/s^2[/tex]

Explanation:

From the question we are told that

       The  mass of the jumbo jet is  [tex]m_j = 100000\ kg[/tex]

        The thrust is  [tex]F_k = 50000 \ N[/tex]

Generally given that the jet has four engines the total thrust is  

        [tex]F_t = 4 * F_k[/tex]

substituting values

       [tex]F_t = 4 * 50000[/tex]

      [tex]F_t = 200000 \ N[/tex]

Generally the acceleration of the is mathematically represented as

         [tex]a = \frac{F_t}{m}[/tex]

substituting values

       [tex]a =2 \frac{N}{kg}[/tex]

Now  

        [tex]N = kg \cdot m/s^2[/tex]

Hence

         [tex]a =2 \frac{kg * \cdot m/s^2}{kg}[/tex]

        [tex]a =2\ m/s^2[/tex]

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Complete Question

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .

Part A. Calculate  the coil's self-inductance.

Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.

Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?

Answer:

Part A  

       [tex]L = 0.000863 \ H[/tex]

Part B  

       [tex]\epsilon = 0.863 \ V[/tex]

Part C

    From terminal a to terminal b

Explanation:

From the question we are told that

      The  number of turns is  [tex]N = 590 \ turns[/tex]

      The cross-sectional area is  [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]

      The  radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]

Generally the coils self -inductance is mathematically represented as

              [tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]

Where [tex]\mu_o[/tex] is the permeability of  free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting values

             [tex]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]

             [tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]

             [tex]L = 0.000863 \ H[/tex]

Considering the Part B

      Initial current is [tex]I_1 = 5.00 \ A[/tex]

      Current at time t is [tex]I_t = 3.0 \ A[/tex]

       The  time taken is  [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]

The self-induced emf is mathematically evaluated as

          [tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]          

=>         [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]

substituting values

             [tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]  

             [tex]\epsilon = 0.863 \ V[/tex]

The direction of the induced emf is  from a to b because according to Lenz's law the induced emf moves in the same direction as the current

This question involves the concepts of the self-inductance, induced emf, and Lenz's Law

A. The coil's self-inductance is "0.863 mH".

B. The self-induced emf in the coil is "0.58 volts".

C. The direction of the induced emf is "from b to a".

A.

The self-inductance of the coil is given by the following formula:

[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]

where,

L = self-inductance = ?

[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²

N = No. of turns = 590

A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²

r = radius = 5 cm = 0.05 m

Therefore,

[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]

L = 0.863 x 10⁻³ H = 0.863 mH

B.

The self-induced emf is given by the following formula:

[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]

where,

E = self-induced emf = ?

ΔI = change in current = 2 A

Δt = change in time = 3 ms = 0.003 s

Therefore,

[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]

E = 0.58 volts

C.

According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.

Learn more about Lenz's Law here:

https://brainly.com/question/12876458?referrer=searchResults