Wire coil with radius r = 14 cm and resistance R

Let us consider a coil of wire with radius r = 14 cm and resistance R = 0.01 Ohm, located in a uniform magnetic field with induction B = 0.2 Tesla.

The plane of the coil makes an angle of 60° with the induction lines.

It is necessary to find the charge flowing through the turn when the magnetic field is turned off.

Answer:

In this problem we are talking about the phenomenon of self-induction, in which a change in the magnetic flux in the coil causes the appearance of a self-induction DC in it.

In our case, when the magnetic field is turned off, the magnetic flux through the coil decreases.

Change of magnetic flux through a coil of wire:

ΔФ = -BSr, where B is the magnetic field induction, S is the cross-sectional area of ​​the coil, r is the radius of the coil.

The angle between the plane of the coil and the induction lines is θ = 60°.

Then the cross-sectional area of ​​the coil is:

S = πr²sinθ = π(0.14)²sin60° ≈ 0.0762 m².

Change in magnetic flux: ΔФ = -0.2 · 0.0762 ≈ -0.01524 Wb.

From the law of self-induction it follows that the DC of self-induction E = -L(dI/dt), where L is the inductance of the coil, I is the current flowing through the coil, t is time.

By definition of inductance, L = ΔФ/I.

Then self-induction DC:

E = -ΔФ/dt = -(L/I)(dI/dt) = -R(dI/dt), where R is the coil resistance.

Thus, the charge flowing through the coil can be found by integrating over time the expression for the DC self-induction:

Q = -∫E dt = -∫(R dI/dt) dt = -R∫dI = -RI + C, where C is the integration constant.

At the initial moment of time, the current in the turn is zero, so the constant C is equal to RI₀, where and₀ - initial current.

Thus, the charge flowing through the coil when the magnetic field is turned off is equal to:

Q = -RI + RI₀ = -0,01 · I + 0,01 · 0 = 0.

Answer: 0 Cl.

Product description

Wire coil

Wire Wrap is a digital product designed for those interested in electricity and magnetism.

Product characteristics

• Radius: 14 cm
• Resistance: 0.01 ohm

Purpose of the product

A coil of wire is used to study the phenomenon of self-induction in electrical circuits. Using this product, you can conduct experiments and demonstrations related to changes in the magnetic flux in the coil and the appearance of self-induction DC.

Product advantages

• High quality workmanship
• Easy to use
• Wide range of applications

The presented product is a coil of wire with a radius of 14 cm and a resistance of 0.01 Ohm, which is used to study the phenomenon of self-induction in electrical circuits. This product allows you to conduct experiments and demonstrations related to changes in the magnetic flux in the coil and the appearance of self-induction DC.

The problem provides information about a coil of wire located in a uniform magnetic field with an induction of 0.2 T; the plane of the coil makes an angle of 60° with the induction lines. It is necessary to find the charge flowing through the turn when the magnetic field is turned off.

To solve the problem, it is necessary to use the law of self-induction, which establishes that a change in the magnetic flux in the coil causes the appearance of a self-induction DC in it. The change in magnetic flux through a coil of wire can be expressed by the formula ΔФ = -BSr, where B is the magnetic field induction, S is the cross-sectional area of ​​the coil, r is the radius of the coil. The cross-sectional area of ​​a coil can be expressed in terms of the radius and the angle between the plane of the coil and the induction lines.

Next, using the definition of inductance and the law of self-induction, one can obtain an expression for the DC self-induction E = -R(dI/dt), where R is the coil resistance, I is the current flowing through the coil, t is time. The charge flowing through the coil when the magnetic field is turned off can be found by integrating over time the expression for the DC self-induction.

So, the charge flowing through the coil when the magnetic field is turned off is equal to -RI + RI0, where R is the resistance of the coil, I is the current flowing through the coil, I0 is the initial current. In this problem, the initial current is zero, so the charge flowing through the coil is 0.

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A coil of wire with radius r = 14 cm and resistance R = 0.01 Ohm is a circular circuit. It is in a uniform magnetic field with induction B = 0.2 Tesla. The plane of the coil makes an angle of 60° with the induction lines.

To solve the problem, it is necessary to use Faraday's law, which establishes that the electromagnetic induction ΔDS in a conductor is equal to the rate of change of the magnetic flux F passing through the surface bounded by the conductor.

The magnetic flux Ф penetrating the surface of a circular coil can be calculated using the formula Ф = B * S * cos(α), where B is the magnetic field induction, S is the surface area limited by the conductor, α is the angle between the direction of magnetic induction and the normal to the surface .

In this case, the surface area of ​​the circular coil is equal to S = π * r^2, angle α = 60° = π/3 radians, since the angle between the direction of magnetic induction and the normal to the surface is 60°. Thus, the magnetic flux Ф penetrating the surface of a circular coil is equal to Ф = B * S * cos(α) = 0.2 * π * (0.14)^2 * cos(π/3) = 0.0254 Wb.

Next, using the formula ?DS of induction, you can calculate ?DS that arises in a wire coil when the magnetic flux changes: E = -dФ/dt, where dФ/dt is the rate of change of the magnetic flux.

When the magnetic field is turned off, the rate of change of the magnetic flux will be maximum and equal to zero until this moment, therefore the resulting ΔDS will be maximum and determined only by the magnitude of the magnetic flux penetrating the surface of the circular coil.

Thus, in this case, ?DS will be equal to E = -dФ/dt = -0.0254 Wb/0 = 0.

Taking into account the fact that ?DS E = -dФ/dt, and charge Q = ∫I dt, where I is the current flowing through the turn at the moment the magnetic field is turned off, we can conclude that the charge flowing through the turn will also be equal zero: Q = ∫I dt = ∫(E/R) dt = E/R * ∫dt = 0.

Answer: the charge flowing through the turn when the magnetic field is turned off is zero.

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