Magnetic Field , Self-inductance & energy Question

In summary, we are given a circular air-core solenoid with a current of 22 A flowing through it. Using Faraday's Law of induction and the equations for magnetic field and self-inductance, we can find the magnetic field at its center to be 1.419 x 10^-4 T, the self-inductance to be 9.027 mH, and the energy stored in the solenoid to be 0.009 J. For the last part, we use Faraday's Law again to find the induced emf in a smaller concentric solenoid placed midway along the length of the larger solenoid. We calculate the emf to be -3.59097 x 10^-
  • #1
Fazza3_uae
51
0

Homework Statement



Given: μ0 = 4 π x10−7 T.m/A.

An air-core circular solenoid is shown in the figure below. A current of 22 A is establishedin the wire which makes up this solenoid.

http://img412.imageshack.us/i/81565602.jpg/


a) What is the magnetic field at its center? [[ Answer in units of T ]]

b) What is its self-inductance? [[ Answer in units of mH ]]

c) How much energy is stored in the solenoid? [[ Answer in units of J ]]




d) A tight circular loop whose diameter is less than the diameter of the solenoid (as shownin the figure below) is concentric with the solenoid and is placed midway along the length of the solenoid. Starting from t = 0, the initial current increases linearly in time until the current doubles in 5 s, and then the current remains constant at 44 A. Find the magnitude of the emf |ε| in the
circular loop at a time t, 0 s < t < 5 s, during the increase of current. [[ Answer in units of V ]]


http://img142.imageshack.us/i/96025878.jpg/


Homework Equations



Faraday's Law V = N d[tex]\phi[/tex]/dt

Self induction = - L di/dt

Energy Stored in magnetic field = 0.5 L I2

[tex]\epsilon[/tex] = -NACos[tex]\theta[/tex] dB/dt

Magnetic field for an ideal selenoid = μ0 i n

The Attempt at a Solution



for part a ) do i have to just plug in the values in the magenetic field equation ??

B = μ0 i n

μ0 is Given
n is given & i is given

For part b ) how to find di/dt ?

for part c ) every thing is given do i have to insert the values and that is it ??

part d ) no idea how to start so any clue or hint will be appreciated.

thx in advance.
 
Last edited:
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  • #2
Try looking in your textbook for the formula that calculates magnetic field & self-inductance of a solenoid.
 
  • #3
hmmm ... i answered the first three parts correctly.

i missed a little informations in those equations above , that's why i could not answer them.

for part a) i used B = μ0 * N * I
i forgot the length L so it will become like this B = μ0 * N * I / L

for part b ) self inductance must equal = μ0 * N^2 * A / L

for part c ) energy stored = 0.5*L*I^2 = 0.5 * B^2 * A * L / μ0

help in the last part d )
 
  • #4
For part D, you have Faraday's law of induction as part of your relevant equations. Just plug in the numbers: phi=? d(phi)/dt=? N=?
 
  • #5
part d)

the emf needed for the small solenoid inside = -n (d[phi]/d[time])

n = 32 turns

d[time] = 5 seconds

d[phi] = B*A

B = μ0 * N * I / L , μ0 is given , N = 323 turns , I = 44 A , L = 0.4 m

A = π r^2 , r = 0.002 m

after finding d[phi] to be = 5.611e-7 T

emf = - (32)*(5.611e-7) /5 = - 3.59097e-6 V

Is it correct now ??
 
  • #6
Fazza3_uae said:
d[phi] = B*A

B = μ0 * N * I / L , μ0 is given , N = 323 turns , I = 44 A , L = 0.4 m

d[phi] is the difference in phi whereas BA is phi itself. In this case, I should be 22 A instead of 44A because it increased by 22 A in five seconds.
 
  • #7
Finally i got the right answer . thanks ideasrule for the help .


what i have done to solve this problem :

The induced emf for the small solenoid is given by Faraday's Law of induction

http://img228.imageshack.us/img228/9731/emf.png


emf = -N2 dφ/dt = -N2 A2dB/dt = -(N2)A2 μo(N1)(dI/dt)/L

N2 = 323 turns
N1 = 32 turns
A2 = π r22 = π 0.0022
dI/dt = [44-22]/[5-0]
L = 0.4 m
μo = 4 π e-7

that is it.
 
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Related to Magnetic Field , Self-inductance & energy Question

1. What is a magnetic field?

A magnetic field is a region in space where a magnetic force can be detected. It is created by moving electric charges, such as electrons, and is represented by lines of force that indicate the direction and strength of the magnetic field.

2. How is a magnetic field produced?

A magnetic field is produced by the movement of electric charges, either through an electric current or by the spin of electrons in an atom. The strength and direction of the magnetic field is determined by the magnitude and direction of the electric current or electron spin.

3. What is self-inductance?

Self-inductance is the property of a circuit or conductor to generate an induced electromotive force (emf) in itself when the current flowing through it changes. This is due to the magnetic field created by the changing current, which induces an emf that opposes the change in current.

4. How is self-inductance measured?

Self-inductance can be measured by using an inductor, which is a passive electronic component designed to store energy in the form of a magnetic field. The inductance of the inductor is a measure of its ability to resist changes in current and is measured in units of henrys (H).

5. What is the relationship between magnetic field and energy?

The energy stored in a magnetic field is directly proportional to the square of the magnetic field strength and the volume of the magnetic field. This means that a stronger magnetic field or a larger volume will result in a greater amount of energy stored in the field.

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