Question

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)

Answer 1

Answer:

N=119.34 turns

Explanation:

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

B= µo*$\frac{I*N}{L}$ 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=$10*10^{-2}$, B= 1.5mT=$1.5*10^{-3} T$  ,I=1A  

µo=$4\pi *10^{-7} \frac{T*m}{A}$

We cleared N of the equation (1):

N=B*L/ µo*I

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

$N=0.1193*10^{3}$

Answer

N=119.34 turns