Finding the current in a solenoid

In summary, to find the maximum current in the coil, we set the current in the solenoid to its maximum value of 0.20A and solve for Io using the equation Isol = Io sin (2Πft). This gives us a value of 140.45A. However, the step where we set t ≈ 0.2387s is incorrect as we have to work in radians, not degrees. Converting 120Hz to radians gives us t ≈ 0.004167s. This gives us a final answer of 140.45A.
  • #1
kamhogo
86
6

Homework Statement


A 40-turn, 4.0-cm-diameter coil with R = 0.40Ω surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The 60 Hz current through the solenoid is I = Io sin (2Πft). What is Io if the maximum current in the coil is 0.20A?

Homework Equations


Ncoil = Nc= 40 turns
Acoil = Ac = 4π*(10^-2) m^2
Rcoil = 0.40Ω
n = Nsol/Lsol = 200 turns / 0.2 m = 1000/m
Bsol = μo*n*Isol
Isol = Io sin (120Πt)
Asol = 2.25π*(10^-4) m^2
φm = Nc*Bsol*Asol (Magnetic flux through coil)
∈coil = φm/dt
Icoil = Nc*∈coil/Rcoil

The Attempt at a Solution


Isol is max when Io sin (2Πft) = Io, i.e. when sin 120Πt = 1, i.e. when t ≈ 0.2387s

From Isol to Bsol
Bsol = 1000*μo*(Io sin (120Πt)

From Bsol to Flux through coil (φm)
φm = 40*(1000*μo*(Io sin (120Πt))*(2.25π*(10^-4) m^2) = 9*(Io sin (120Πt)

From φm to ∈coil
∈coil = (9*(Io sin (120Πt))/dt= 9*Io*120Π* cos (120Πt) = 3392.92*Io*cos (120Πt)

From ∈coil to Icoil
Icoil = (40*3392.92*Io*cos (120Πt))/0.40Ω = 339292.0066 *Io*cos (120Πt) = 0.20 A
5.8946*(10^-7) = Io*cos (120Πt); t ≈ 0.2387 s ⇒ Io = 140.45A

Is this correct?
 
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  • #2
kamhogo said:

Homework Statement


A 40-turn, 4.0-cm-diameter coil with R = 0.40Ω surrounds a 3.0-cm-diameter solenoid. The solenoid is 20 cm long and has 200 turns. The 60 Hz current through the solenoid is I = Io sin (2Πft). What is Io if the maximum current in the coil is 0.20A?

Homework Equations


Ncoil = Nc= 40 turns
Acoil = Ac = 4π*(10^-2) m^2
Rcoil = 0.40Ω
n = Nsol/Lsol = 200 turns / 0.2 m = 1000/m
Bsol = μo*n*Isol
Isol = Io sin (120Πt)
Asol = 2.25π*(10^-4) m^2
φm = Nc*Bsol*Asol (Magnetic flux through coil)
∈coil = φm/dt
Icoil = Nc*∈coil/Rcoil

The Attempt at a Solution


Isol is max when Io sin (2Πft) = Io, i.e. when sin 120Πt = 1, i.e. when t ≈ 0.2387s
That step is incorrect. You have to work in radians.
 
  • #3
nrqed said:
That step is incorrect. You have to work in radians.
So 120*pi*t = 1.570796327
t = 0.004167 s ...??
And why in radians?
 
  • #4
kamhogo said:
So 120*pi*t = 0.5*pi
t = (1/240) s ...??
And why in radians?

But then in the last step I get :

(5.8946*10^(-7)) / cos (0.5*pi) = infinity...
 

FAQ: Finding the current in a solenoid

1. What is a solenoid?

A solenoid is a type of electromagnet that consists of a coil of wire wrapped around a cylindrical core. When an electric current flows through the wire, it creates a magnetic field that can be used for various applications.

2. How do you find the current in a solenoid?

The current in a solenoid can be found using the equation I = NΦ/t, where I is the current in amperes, N is the number of turns in the coil, Φ is the magnetic flux through the solenoid, and t is the time in seconds.

3. What factors affect the current in a solenoid?

The current in a solenoid can be affected by several factors, including the number of turns in the coil, the strength of the magnetic field, the length and diameter of the solenoid, and the material of the core. Additionally, the resistance of the wire and the voltage applied to the solenoid can also affect the current.

4. How does the current in a solenoid change with time?

If the voltage applied to a solenoid remains constant, the current in the solenoid will also remain constant. However, if the voltage is changed, the current will change accordingly. Additionally, if the solenoid is connected to a power source with a varying voltage, the current will also vary over time.

5. How is the current in a solenoid related to its magnetic field?

The current in a solenoid is directly related to its magnetic field. When an electric current flows through the wire, it creates a magnetic field within and around the solenoid. The strength of this magnetic field is directly proportional to the current in the solenoid, with a greater current resulting in a stronger magnetic field.

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