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------------------------------------------------ A solenoid is constructed from 5000 turns of wire onto a form of length 6.28 cm and diameter 4.513 cm. The solenoid is connected to a battery so that the current increases from 0 A to 4.00 mA in a time of 8.00 ms.

1. Calculate B inside the coil at 8.00 ms, in μT.

2. What is the inductance of the solenoid, in henrys?

3. What is the absolute value of EMF induced in the inductor, in mV?

4. What is the maximum energy stored in the inductor, in μJ?

5. Determine the energy density inside the inductor, in mJ/m³.

6. Insert an iron core with $K_m = 1000$ into the bore of the solenoid, completely filling it. Calculate the new self-inductance (L) in henrys.

7. What voltage (V) would be induced if the current increases from 0 to 4.00 mA in a time of 8.00 ms, in volts?

8. How much energy is stored in the inductor, in mJ?

9. Insert the original coil without iron into another coil of length 6.28 cm and diameter 9.026 cm with 2500 turns. Calculate the mutual inductance (M) between the two coils, in henrys.

10. If the inner solenoid is connected to a battery so that the current increases from 0 A to 4.00 mA in a time of 8.00 ms, calculate the absolute value of the voltage V induced in the secondary coil at a time of 8.00 ms, in mV.

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).

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