# Finding the current in a solenoid

• kamhogo

## 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?

## 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.

That step is incorrect. You have to work in radians.

So 120*pi*t = 1.570796327
t = 0.004167 s ...??