Answer :
B ≈ 0.032 T or 32 μT (microteslas).
L ≈ 0.636 H (henries).
EMF ≈ 0.318 V or 318 mV (millivolts).
E ≈ 0.00127 J or 1.27 μJ (microjoules).
U ≈ 0.00159 J/m³ or 1.59 mJ/m³ (millijoules per cubic meter).
L' = (1000 * 4π × 10^-7 H/m * N² * A) / l
V ≈ 0.318 V or 318 mV (millivolts).
E ≈ 0.00127 J or 1.27 mJ (millijoules).
M ≈ 0.255 H or 255 mH (millihenries).
V ≈ 0.127 V or 127 mV (millivolts).
1. To calculate the magnetic field (B) inside the solenoid at 8.00 ms, we can use the formula:
B = μ₀ * (N / L) * I
where μ₀ is the permeability of free space (4π * 10^-7 T·m/A), N is the number of turns (5000), L is the length of the solenoid (6.28 cm or 0.0628 m), and I is the current (4.00 mA or 0.004 A).
Plugging in these values, we get:
B = (4π * 10^-7) * (5000 / 0.0628) * 0.004
B ≈ 0.032 T or 32 μT (microteslas).
2. The inductance (L) of the solenoid can be calculated using the formula:
L = (μ₀ * N² * A) / l
where A is the cross-sectional area of the solenoid (π * r²), r is the radius of the solenoid (half of the diameter), and l is the length of the solenoid.
Plugging in the values, we get:
A = π * (4.513 cm / 2)^2 ≈ 15.964 cm² or 0.0015964 m²
L = (4π * 10^-7) * (5000²) * (0.0015964) / 0.0628
L ≈ 0.636 H (henries).
3. The absolute value of the electromotive force (EMF) induced in the inductor can be calculated using the formula:
EMF = L * (ΔI / Δt)
where ΔI is the change in current (4.00 mA or 0.004 A) and Δt is the change in time (8.00 ms or 0.008 s).
Plugging in these values, we get:
EMF = 0.636 * (0.004 / 0.008)
EMF ≈ 0.318 V or 318 mV (millivolts).
4. The maximum energy stored in the inductor can be calculated using the formula:
E = (1/2) * L * I²
where I is the maximum current (0.004 A).
Plugging in the values, we get:
E = (1/2) * 0.636 * (0.004)²
E ≈ 0.00127 J or 1.27 μJ (microjoules).
5. The energy density inside the inductor can be calculated using the formula:
U = (1/2) * B² / μ₀
where B is the magnetic field inside the solenoid at 8.00 ms (0.032 T or 32 μT).
Plugging in the value, we get:
U = (1/2) * (0.032)² / (4π * 10^-7)
U ≈ 0.00159 J/m³ or 1.59 mJ/m³ (millijoules per cubic meter).
6. To calculate the new self-inductance (L) of a solenoid after inserting an iron core with a relative permeability (Km) of 1000, we need to consider the effect of the core material on the magnetic field within the solenoid.
The self-inductance of a solenoid is given by the equation:
L = (μ₀ * N² * A) / l
where L is the self-inductance, μ₀ is the permeability of free space (4π × 10^-7 H/m), N is the number of turns in thesolenoid, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
When an iron core with a high relative permeability is inserted into the solenoid, the effective permeability of the solenoid increases. In this case, Km = 1000.
The relative permeability (K) is given by:
K = Km * μ₀
Substituting the value of Km, we have:
K = 1000 * 4π × 10^-7 H/m
Now, we can calculate the new self-inductance (L'):
L' = (K * N² * A) / l
Substituting the value of K, we have:
L' = (1000 * 4π × 10^-7 H/m * N² * A) / l
7. The voltage (V) induced when the current increases from 0 to 4.00 mA in 8.00 ms can be calculated using the formula:
V = L * (ΔI / Δt)
where L is the inductance (0.636 H), ΔI is the change in current (0.004 A), and Δt is the change in time (0.008 s).
Plugging in the values, we get:
V = 0.636 * (0.004 / 0.008)
V ≈ 0.318 V or 318 mV (millivolts).
8. The energy stored in the inductor can be calculated using the formula:
E = (1/2) * L * I²
where L is the inductance (0.636 H) and I is the current (0.004 A).
Plugging in the values, we get:
E = (1/2) * 0.636 * (0.004)²
E ≈ 0.00127 J or 1.27 mJ (millijoules).
9. To calculate the mutual inductance (M) between the two coils, we can use the formula:
M = (μ₀ * N₁ * N₂ * A) / l
where N₁ is the number of turns in the first coil (5000), N₂ is the number of turns in the second coil (2500), A is the cross-sectional area of the coils (π * r²), r is the radius of the coils (half of the diameter), and l is the length of the coils.
Plugging in the values, we get:
A = π * (9.026 cm / 2)^2 ≈ 64.077 cm² or 0.0064077 m²
M = (4π * 10^-7) * (5000) * (2500) * (0.0064077) / 0.0628
M ≈ 0.255 H or 255 mH (millihenries).
10. The absolute value of the voltage (V) induced in the secondary coil at 8.00 ms can be calculated using the formula:
V = M * (ΔI / Δt)
where M is the mutual inductance (0.255 H), ΔI is the change in current (0.004 A), and Δt is the change in time (0.008 s).
Plugging in the values, we get:
V = 0.255 * (0.004 / 0.008)
V ≈ 0.127 V or 127 mV (millivolts).
Know more about magnetic field here,
https://brainly.com/question/14848188
#SPJ11
