How many turns should a 10-cm long ideal solenoid have if it is to generate a 1.5-mT magnetic field when 1.0 A of current runs through it? (μ0 = 4???? × 10-7 T · m/A)

Respuesta :

Answer:

N=119.34 turns

Explanation:

The magnetic field of a solenoid is calculated using the formula:

B= µo*[tex]\frac{I*N}{L}[/tex] Equation 1

Where:

B: magnetic field in Teslas (T)

µo: free space permeability in T*m/A

I= Intensity of the current flowing through the conductor in ampere (A)

N= number of turns

L= solenoid length in meters (m)

Data of the problem:

L=10cm=[tex]10*10^{-2}[/tex], B= 1.5mT=[tex]1.5*10^{-3} T[/tex]  ,I=1A  

µo=[tex]4\pi *10^{-7} \frac{T*m}{A}[/tex]

We cleared N of the equation (1):

N=B*L/ µo*I

N= [tex](1.5*10^{-3} *10*10^{-2} )/(4\pi *10^{-7} *1[/tex]

[tex]N=0.1193*10^{3}[/tex]

Answer

N=119.34 turns