Typical problems with solutions

Typical tasks with solutions and for independent solution

On the topic “Ferromagnets and their properties”

Typical problems with solutions

Problem 1. A solenoid with a core (magnetic permeability of the core material μ = 100) has a dense single-layer winding with a wire 0.2 mm in diameter, and a current of 0.1 A flows through it. The length of the solenoid is 20 cm, the diameter is 5 cm. Find the energy of the magnetic field of the solenoid .

Given: d = 0.2 mm I = 0.1 A D = 5 cm l = 20 cm μ = 100 SI 2×10 -4 m 5×10 -2 m 0.2 m
w-?

Decision:

The energy of the magnetic field of the solenoid is:

,

where – solenoid inductance,

n is the number of turns per 1 m of the solenoid.

When tightly wound .

Solenoid cross-sectional area .

Then

Answer: W = 16.43 mJ.

Task 2. A single-layer solenoid with a core (magnetic permeability of the core material μ = 100) 20 cm long and 4 cm in diameter is tightly wound with copper wire 0.1 mm in diameter. For 0.1 s, the current in it decreases uniformly from 5 A to 0. Determine the EMF of self-induction in the solenoid.

Given: d u003d 0.1 mm I 1 u003d 5 A I 2 u003d 0 D u003d 4 cm l u003d 20 cm μ u003d 100 SI 10 -4 m 4×10 -2 m 0.2 m
ε-?

Decision:

EMF of self-induction that occurs when the current ΔI in the solenoid changes during the time Δt:

,

where – solenoid inductance,

n is the number of turns per 1 m of the solenoid.

When tightly wound .

Answer: ε= 157.8 V.

Task 3. A solenoid with a length of 20 cm and a diameter of 4 cm has a dense three-layer winding made of a wire with a diameter of 0.1 mm. A current of 0.1 A flows through the solenoid winding. The dependence B u003d f (H) for the core material is given in Figure 1. Determine the field strength and induction in the solenoid, the magnetic permeability of the core, the solenoid inductance, the energy and volume energy density of the solenoid field.

Picture 1.

Given: d = 0.1 mm I = 0.1 A D = 4 cm l = 20 cm k = 3 SI 10 -4 m 4×10 -2 m 0.2 m
H, B, L, μ, W, w -?

Decision:

The field inside the solenoid can be considered uniform. In this case, the field strength is:

H = I n,

where n is the number of turns per unit length of the solenoid:

.

Here k is the number of winding layers, d is the wire diameter.

Then

.

According to the graph B = f(H) (Fig. 1), we find that the intensity of 3000 A/m corresponds to an induction of 1.7 T.

Using the relationship between induction and intensity, we determine the magnetic permeability:

B = μμ 0 H,

.

Solenoid inductance

.

Taking into account the fact that and , we get:

Volumetric energy density of the magnetic field:

Solenoid magnetic field energy

.

Answer: H= 3000 A/m; , μ=451; L=128 H; w=2.55 kJ/m 3 ; W=0.64 J.

Problem 4. An insulated ring of the same diameter is put on the solenoid (see the condition and solution of the previous problem). Determine the EMF of induction in the ring and the EMF of self-induction in the solenoid, if in 0.01 s the current in its winding decreases uniformly to zero.

Given: V u003d 1.7 T I 1 u003d 0.1 A I 2 u003d 0 D u003d 4 cm L u003d 128 H Δt u003d 0.01 c to u003d 3 SI 4×10 -2 m
ε i , ε si -?

Decision:

According to the condition of the problem, during the time Δt = 0.01 s, the current in the solenoid winding decreases uniformly from 0.1 A to 0, so the magnetic flux penetrating the ring area , will decrease from Ф 1 = BS to Ф 2 = 0. The induction emf that occurs in the ring,

.

EMF of self-induction ε si that occurs in the solenoid when the current is turned off in it, .

Then

Answer: ε i = 0.21 V; εsi = 1280 V.

Task 5. A current of 5A flows through the solenoid. The length of the solenoid is 1 m, the number of turns is 500. An iron core is inserted into the solenoid. Find the magnetization and volume energy density of the magnetic field of the solenoid. The dependence B = f(H) is given in Figure 1 (task 3).

Given: I = 5 A l = 1 m N = 500 SI
J, w-?

Decision:

Magnetization is determined by the ratio of the magnetic moment to the volume of the magnet and is related to the magnetic field strength by the relation:

J = χ H,

where χ is the magnetic susceptibility of the medium.

The solenoid field can be considered uniform. In this case, the field strength is calculated by the formula:

H = I n.

Then .

The relationship between the magnetic susceptibility χ and the magnetic permeability μ of the medium is expressed by the formula:

χ = μ – 1.

Let’s determine the magnetic field strength:

.

According to the graph (Fig. 1 to problem 3), we find that the intensity H = 2500 A/m corresponds to the magnetic field induction B = 1.6 T. Using the relation B = μμ 0 H, we determine μ:

.

χ u003d μ – 1 u003d 509 – 1 u003d 508.

J u003d χ H u003d 508 2500 u003d 12.7 A / m.

The volumetric energy density of the magnetic field of the solenoid is calculated by the formula:

.

Answer: J=12.7 A/m; w u003d 2.55 kJ / m 3 .

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