Respuesta :
The magnitude of the magnetic field in the central area inside the solenoid, in T is 0.0267 T
Magnetic field inside solenoid
The magnetic field inside the central area of the solenoid is given by B = μ₀ni where
- μ₀ = permeability of free space = 4π × 10⁻⁷ Tm/A,
- n = number of turns per unit length = 3,170 turns/m and
- i = current in solenoid = 6.7 A
Since B = μ₀ni
Substituting the values of the variables into the equation, we have
B = μ₀ni
B = 4π × 10⁻⁷ Tm/A × 3,170 turns/m × 6.7 A
B = 4π × 10⁻⁷ Tm/A × 21239 A-turns/m
B = 84956π × 10⁻⁷ T
B = 266897.15 × 10⁻⁷ T
B = 0.026689715 T
B ≅ 0.0267 T
So, the magnitude of the magnetic field in the central area inside the solenoid, in T is 0.0267 T
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The magnitude of the magnetic field in the central area inside the considered solenoid (assuming in vaccum), is approx 2.67 × 10^{-2} T
What is the measurement of the magnetic field inside a long solenoid?
Magnetic field = permeability constant x relative permeability x turn density x current
B = [tex]k\mu_0nI = \mu nI[/tex]
where turn density is calculated as:
n = N/L
Where N represents the number of turns of the wire and
L represents the length of the solenoid.
Assuming that the solenoid is placed in vaccum, we have:
[tex]\mu = k\mu_0 = \mu_0 = 4\pi \times 10^{-7} \: \rm T m/A[/tex]
And it is given that:
- n = 3174 turns/m
- I = 6.7 amperes
Thus, we get:
[tex]B = \mu n I = 4\pi \times 10^{-7} \times 3170 \times 6.7 \approx 2.67 \times 10^5 \times 10^{-7} \\\\B = 2.67 \times 10^{5 + (-7)}B = 2.67 \times 10^{-2} \: \tm T[/tex]
Thus, the magnitude of the magnetic field in the central area inside the considered solenoid (assuming in vaccum), is approx 2.67 × 10^{-2} T
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