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Magnetic Field , Selfinductance & energy Question 
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#1
Nov2809, 03:30 AM

P: 51

1. The problem statement, all variables and given/known data
Given: μ_{0} = 4 π x10^{−7 }T.m/A. An aircore 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 selfinductance? [[ 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 I^{2} [tex]\epsilon[/tex] = NACos[tex]\theta[/tex] dB/dt Magnetic field for an ideal selenoid = μ_{0} 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 = μ_{0} i n μ_{0} 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. 


#2
Nov2809, 03:39 AM

HW Helper
P: 2,323

Try looking in your textbook for the formula that calculates magnetic field & selfinductance of a solenoid.



#3
Nov2809, 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 ) 


#4
Nov2809, 05:20 AM

HW Helper
P: 2,323

Magnetic Field , Selfinductance & 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=?



#5
Nov2809, 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 = 5.611e7 T emf =  (32)*(5.611e7) /5 =  3.59097e6 V Is it correct now ?? 


#6
Nov2809, 11:04 AM

HW Helper
P: 2,323




#7
Nov2809, 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 = N_{2} dφ/dt = N_{2} A_{2}dB/dt = (N_{2})A_{2} μ_{o}(N_{1})(dI/dt)/L N2 = 323 turns N1 = 32 turns A2 = π r_{2}^{2} = π 0.002^{2} dI/dt = [4422]/[50] L = 0.4 m μ_{o} = 4 π e^{7} that is it. 


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