A coil with 150 turns , a radius of 5.0 cm , and a resistance of 12 ohms surrounds a solenoid with 230 turns/cm,and a radius of 4.5 cm. The current in the solenoid changes at a constant rate from 0 to 2 A in 0.10 sec. Calculate the magnitude and direction of the induced current in the coil.
To calculate the magnitude and direction of the induced current in the coil, we can use Faraday's law of electromagnetic induction, which states that the magnitude of the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through the coil.
The magnetic flux through the coil can be calculated using the formula:
Φ = BA
Where:
Φ = Magnetic flux
B = Magnetic field
A = Area
The magnetic field inside a solenoid is given by the formula:
B = μ₀ * N * I
Where:
B = Magnetic field
μ₀ = Permeability of free space = 4π × 10⁻⁷ Tm/A
N = Number of turns per unit length (turns/cm)
I = Current
Let's calculate the magnetic field inside the solenoid:
B = (4π × 10⁻⁷ Tm/A) * (230 turns/cm) * (2 A)
B = 0.00276 T
Now, let's calculate the magnetic flux through the coil:
Φ = (0.00276 T) * (π * (0.05 m)²)
Φ = 6.89 × 10⁻⁶ Wb
The induced electromotive force (EMF) can be calculated using the formula:
EMF = -N * dΦ/dt
Where:
EMF = Electromotive force
N = Number of turns in the coil
dΦ/dt = Rate of change of magnetic flux with respect to time
Since the current in the solenoid changes at a constant rate from 0 to 2 A in 0.10 sec, we can calculate the rate of change of magnetic flux:
dΦ/dt = (6.89 × 10⁻⁶ Wb) / (0.10 s)
dΦ/dt = 6.89 × 10⁻⁵ Wb/s
Now, let's calculate the induced EMF:
EMF = -(150 turns) * (6.89 × 10⁻⁵ Wb/s)
EMF = -0.0103 V
Finally, to calculate the magnitude and direction of the induced current in the coil, we can use Ohm's Law:
EMF = I * R
Where:
I = Induced current
R = Resistance
Rearranging the equation to solve for the induced current:
I = EMF / R
I = (-0.0103 V) / (12 Ω)
I ≈ -8.58 × 10⁻⁴ A
Therefore, the magnitude of the induced current in the coil is approximately 8.58 × 10⁻⁴ A, and the direction is opposite to the direction of the changing current in the solenoid.
To calculate the magnitude and direction of the induced current in the coil, we can use Faraday's law of electromagnetic induction. Faraday's law states that the induced emf (electromotive force) is equal to the negative rate of change of magnetic flux through a surface.
1. First, let's calculate the change in magnetic flux through the coil. The magnetic flux is given by the formula:
Φ = B * A
where Φ is the magnetic flux, B is the magnetic field, and A is the area.
2. The magnetic field (B) inside a solenoid is given by:
B = μ₀ * n * I
where μ₀ is the permeability of free space (4π x 10^-7 T m/A), n is the number of turns per unit length (230 turns/cm = 23000 turns/m), and I is the current in the solenoid.
3. The area (A) of the coil is given by:
A = π * r²
where r is the radius of the coil.
4. The change in flux (ΔΦ) is given by:
ΔΦ = Φ₂ - Φ₁
where Φ₂ is the final flux and Φ₁ is the initial flux.
5. The initial flux (Φ₁) is the flux when the current is 0 A, so we can substitute the values into the above formulas to find it.
6. The final flux (Φ₂) is the flux when the current is 2 A, so we can substitute the values into the above formulas to find it.
7. The induced emf (ε) is given by:
ε = -dΦ/dt
where dt is the change in time.
8. The induced current (Iind) in the coil is given by Ohm's law:
Iind = ε / R
where R is the resistance of the coil.
Let's calculate:
Step 1: Calculating the initial flux (Φ₁)
B₁ = μ₀ * n * I₁
= 4π x 10^-7 T m/A * 23000 turns/m * 0 A
= 0 T
A = π * r²
= π * (5.0 cm)^2
= 78.54 cm²
= 0.007854 m²
Φ₁ = B₁ * A
= 0 T * 0.007854 m²
= 0 Wb (Weber)
Step 2: Calculating the final flux (Φ₂)
B₂ = μ₀ * n * I₂
= 4π x 10^-7 T m/A * 23000 turns/m * 2 A
= 0.0584 T
Φ₂ = B₂ * A
= 0.0584 T * 0.007854 m²
= 4.59 x 10^-4 Wb
Step 3: Calculating the change in flux (ΔΦ)
ΔΦ = Φ₂ - Φ₁
= 4.59 x 10^-4 Wb - 0 Wb
= 4.59 x 10^-4 Wb
Step 4: Calculating the induced emf (ε)
ε = -dΦ/dt
= -ΔΦ / dt
= -4.59 x 10^-4 Wb / 0.10 s
= -4.59 x 10^-3 V
Step 5: Calculating the induced current (Iind)
Iind = ε / R
= (- 4.59 x 10^-3 V) / 12 Ω
≈ -3.83 x 10^-4 A
The magnitude of the induced current in the coil is approximately 3.83 x 10^-4 A (or 0.383 mA), and the direction is opposite to the change in magnetic field (or opposite to the direction of the current in the solenoid).