A 500 turn solenoid with a length of 20 cm and a radius of 1.5 cm carries a current of 2.0 A. A second coil of four turns is wrapped tightly about this solenoid so that it can be considered to have the same radius as the solenoid. Find the following when the current in the solenoid increases to 5.0 A in a period of 0.90 s.
(a) the change in the magnetic flux through the coil
T·m2
(b) the magnitude of the average induced emf in the coil
V
To find the change in the magnetic flux through the coil and the magnitude of the average induced emf in the coil, we can use Faraday's Law of electromagnetic induction.
(a) The change in magnetic flux through the coil can be calculated using the formula:
ΔΦ = NΔB
Where ΔΦ is the change in magnetic flux, N is the number of turns in the coil, and ΔB is the change in magnetic field.
Given:
Number of turns in the coil (N) = 4 (given)
Change in current (ΔI) = 5.0 A - 2.0 A = 3.0 A
Change in time (Δt) = 0.90 s
Radius of the solenoid (r) = 1.5 cm = 0.015 m
We can find ΔB using the formula:
ΔB = μ₀ΔI/Δt
Where μ₀ is the permeability of free space.
The value of μ₀ is 4π × 10^(-7) T·m/A.
Substituting the given values:
ΔB = (4π × 10^(-7) T·m/A) × (3.0 A) / (0.90 s)
≈ 4.1888 × 10^(-6) T
Now, we can calculate the change in magnetic flux:
ΔΦ = NΔB
= 4 (turns) × 4.1888 × 10^(-6) T
≈ 1.675 × 10^(-5) T·m²
Therefore, the change in the magnetic flux through the coil is approximately 1.675 × 10^(-5) T·m².
(b) The magnitude of the average induced emf (ε) in the coil can be calculated using the formula:
ε = -N (ΔΦ/Δt)
Substituting the known values:
ε = -4 (turns) × [(1.675 × 10^(-5) T·m²) / (0.90 s)]
≈ -7.389 × 10^(-5) V
Note: The negative sign indicates that the direction of the induced emf is opposite to the change in the current.
Therefore, the magnitude of the average induced emf in the coil is approximately 7.389 × 10^(-5) V.
To answer this question, we need to calculate the change in magnetic flux and the average induced emf in the coil.
(a) The change in magnetic flux (ΔΦ) through the coil can be calculated using Faraday's Law of electromagnetic induction, which states that the induced emf (ε) is equal to the rate of change of magnetic flux.
ΔΦ = -NΔB·A
Where:
- ΔΦ is the change in magnetic flux
- N is the number of turns in the second coil
- ΔB is the change in magnetic field strength
- A is the area of the coil
First, we need to find the change in magnetic field strength (ΔB) by using Ampere's Law. Ampere's Law states that the magnetic field strength (B) inside a solenoid is directly proportional to the current (I) and the number of turns per unit length (n) of the solenoid.
B = μ₀nI
Where:
- B is the magnetic field strength
- μ₀ is the permeability of free space (approximately 4π x 10^-7 T·m/A)
- n is the number of turns per unit length (n = N/L, where N is the total number of turns and L is the length of the solenoid)
- I is the current in the solenoid
Substituting this equation into the first equation:
ΔΦ = -NLΔ(B/A)
Now, we need to calculate the change in magnetic field strength (ΔB) by subtracting the initial magnetic field strength from the final magnetic field strength.
ΔB = Bf - Bi
The initial magnetic field strength (Bi) is calculated using the initial current (Ii) in the solenoid:
Bi = μ₀nIi
The final magnetic field strength (Bf) is calculated using the final current (If) in the solenoid:
Bf = μ₀nIf
Substituting these values into the equation for ΔΦ:
ΔΦ = -NL(μ₀nIf - μ₀nIi) = -μ₀N²L(nIf - nIi)
Plug in the given values:
N = 4 (number of turns in the second coil)
L = 0.20 m (length of the solenoid)
n = N/L = 4/0.20 = 20 turns/m (number of turns per unit length)
Ii = 2.0 A (initial current in the solenoid)
If = 5.0 A (final current in the solenoid)
μ₀ = 4π x 10^-7 T·m/A (permeability of free space)
ΔΦ = - (4π x 10^-7 T·m/A)(4²)(0.20 m)((20 turns/m)(5.0 A - 2.0 A))
(b) The magnitude of the average induced emf (ε) in the coil can be calculated using Faraday's Law:
ε = -N(dΦ/dt)
Where:
- N is the number of turns in the second coil
- dΦ/dt is the rate of change of magnetic flux
Taking the derivative of ΔΦ with respect to time (t):
ΔΦ/Δt = (4π x 10^-7 T·m/A)(4²)(0.20 m)((20 turns/m)(5.0 A - 2.0 A))/(0.90 s)
This calculation will give you the average induced emf (ε) in volts (V).
Please note that in this calculation, we assume that the magnetic field induced by the solenoid's own magnetic field is negligible compared to the magnetic field induced by the changing current in the solenoid.