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Magnetic Field , Self-inductance & energy Question

by Fazza3_uae
Tags: energy, field, magnetic, selfinductance
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Fazza3_uae
#1
Nov28-09, 03:30 AM
P: 51
1. The problem statement, all variables and given/known data

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/


2. Relevant 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

3. 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.
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ideasrule
#2
Nov28-09, 03:39 AM
HW Helper
ideasrule's Avatar
P: 2,322
Try looking in your textbook for the formula that calculates magnetic field & self-inductance of a solenoid.
Fazza3_uae
#3
Nov28-09, 04:34 AM
P: 51
hmmm ... i answered the first three parts correctly.

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

for part a) i used B = μ0 * N * I
i forgot the lenght 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 )

ideasrule
#4
Nov28-09, 05:20 AM
HW Helper
ideasrule's Avatar
P: 2,322
Magnetic Field , Self-inductance & energy Question

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=?
Fazza3_uae
#5
Nov28-09, 06:37 AM
P: 51
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 ??
ideasrule
#6
Nov28-09, 11:04 AM
HW Helper
ideasrule's Avatar
P: 2,322
Quote Quote by Fazza3_uae View Post
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.
Fazza3_uae
#7
Nov28-09, 11:56 AM
P: 51
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




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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