magnetism?
2015-09-02 3:43 am
A solenoid that is 62 cm long produces a magnetic field of 1.3 T within its core when it carries a current of 8.4 A. How many turns of wire are contained in this solenoid?
回答 (2)
2015-09-02 3:50 am
If I remember correctly the equation is B = mu n I.
B is the magnetic field in Tesla.
mu is the magnetic permeability of free space = 4pi 10^-7 T m/A
** Im assuming the core is air or vacuum.
I is the current in amperes.
n is the turn density in turns/m.
Solve for n
n = B / (mu I)
n = 1.3 T / (4 pi * 10^-7 T m / A * 8.4 A)
n = 123,156 turns / m
And since the solenoid is 0.62 meters long...
N = 76,356 turns.
B is the magnetic field in Tesla.
mu is the magnetic permeability of free space = 4pi 10^-7 T m/A
** Im assuming the core is air or vacuum.
I is the current in amperes.
n is the turn density in turns/m.
Solve for n
n = B / (mu I)
n = 1.3 T / (4 pi * 10^-7 T m / A * 8.4 A)
n = 123,156 turns / m
And since the solenoid is 0.62 meters long...
N = 76,356 turns.
2015-09-02 4:54 am
Magnetic field B in Tesla = μ*I*n/ℓ
where
μ = 1,256*10^-6 T*m/Amp
I = currents in Amps
n = number of turns
ℓ = coil lenght in m
so :
1.3 = 1.256*10^-6*8.4*n/0.62
n = 0.806*10^6/(1.256*8.4) = 76395 turns
Flux Φ = n*B*A = (μ*I*n/ℓ)*n*A = μ*I*n^2*A/ℓ
A being the coil's cross section in m^2 assumed to be 2*10^-4 m^2
inductance L = Φ/I = 1.3*7.64*10^4*2*10^-4/8.4 = 2.36 Henry approx.
where
μ = 1,256*10^-6 T*m/Amp
I = currents in Amps
n = number of turns
ℓ = coil lenght in m
so :
1.3 = 1.256*10^-6*8.4*n/0.62
n = 0.806*10^6/(1.256*8.4) = 76395 turns
Flux Φ = n*B*A = (μ*I*n/ℓ)*n*A = μ*I*n^2*A/ℓ
A being the coil's cross section in m^2 assumed to be 2*10^-4 m^2
inductance L = Φ/I = 1.3*7.64*10^4*2*10^-4/8.4 = 2.36 Henry approx.
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