Problem 29 - College Physics

Question

Two coils are wound around the same cylindrical form, like the coils in Example \(21.8 .\) When the current in the first coil is decreasing at a rate of \(0.242 \mathrm{A} / \mathrm{s},\) the induced emf in the second coil has magnitude 1.65 \(\mathrm{mV}\) . (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the average magnetic flux through each turn when the current in the first coil equals 1.20 \(\mathrm{A} ?(\mathrm{c})\) If the current in the second coil increases at a rate of \(0.360 \mathrm{A} / \mathrm{s},\) what is the magnitude of the induced emf in the first coil?

Step-by-Step Solution

Answer

(a) \( M \approx 6.82 \times 10^{-3} \,\mathrm{H} \); (b) \( \Phi_{21} \approx 3.27 \times 10^{-4} \,\mathrm{Wb} \); (c) \( \mathcal{E}_{12} \approx 2.46 \,\mathrm{mV} \).

Step 1: Understanding the Given Data for Part (a)

We are given that the rate of change of current in the first coil \( \frac{dI_1}{dt} = -0.242 \,\mathrm{A/s} \) and the induced emf in the second coil \( \mathcal{E}_{21} = 1.65 \,\mathrm{mV} = 1.65 \times 10^{-3} \,\mathrm{V} \). Using the mutual inductance formula \( \mathcal{E}_{21} = M \frac{dI_1}{dt} \), we can solve for \( M \).

Step 2: Calculating the Mutual Inductance for Part (a)

Rearrange the formula to solve for \( M \):\[ M = \frac{\mathcal{E}_{21}}{\frac{dI_1}{dt}} = \frac{1.65 \times 10^{-3}}{-0.242} \,\mathrm{H}. \]Calculate \( M \) using these values.

Step 3: Calculation Result for Mutual Inductance

Substituting the values, we get:\[ M = \frac{1.65 \times 10^{-3}}{0.242} \approx 6.82 \times 10^{-3} \,\mathrm{H}. \] This is the mutual inductance of the pair of coils.

Step 4: Understanding the Given Data for Part (b)

We know \( N_2 = 25 \), the number of turns in the second coil. We need to find the average magnetic flux \( \Phi_{21} \) through each turn when the current in the first coil is \( 1.20 \,\mathrm{A} \). Use the formula \( \Phi_{21} = \frac{M I_1}{N_2} \) where \( I_1 = 1.20 \,\mathrm{A} \).

Step 5: Calculating the Magnetic Flux for Part (b)

Using the mutual inductance \( M = 6.82 \times 10^{-3} \,\mathrm{H} \), we find:\[ \Phi_{21} = \frac{6.82 \times 10^{-3} \times 1.20}{25}. \] Calculate \( \Phi_{21} \) using this expression.

Step 6: Calculation Result for Magnetic Flux

Substituting the values, we get:\[ \Phi_{21} = \frac{6.82 \times 10^{-3} \times 1.20}{25} \approx 3.27 \times 10^{-4} \,\mathrm{Wb} \] per turn.

Step 7: Understanding the Given Data for Part (c)

The rate of change of current in the second coil is \( \frac{dI_2}{dt} = 0.360 \,\mathrm{A/s} \). We need to find the induced emf in the first coil \( \mathcal{E}_{12} \) using \( \mathcal{E}_{12} = M \frac{dI_2}{dt} \).

Step 8: Calculating the Induced EMF for Part (c)

Using the mutual inductance \( M = 6.82 \times 10^{-3} \,\mathrm{H} \), we find:\[ \mathcal{E}_{12} = 6.82 \times 10^{-3} \times 0.360. \]Calculate \( \mathcal{E}_{12} \) using these values.

Step 9: Calculation Result for Induced EMF

Substituting the values, we get:\[ \mathcal{E}_{12} = 6.82 \times 10^{-3} \times 0.360 \approx 2.46 \times 10^{-3} \,\mathrm{V} = 2.46 \,\mathrm{mV}. \]This is the magnitude of the induced emf in the first coil.

Key Concepts

Magnetic Flux

Magnetic flux is a measure of the amount of magnetic field passing through an area. It is an essential concept in understanding how magnetic fields interact with materials, like coils in this exercise. Magnetic flux (\( \Phi \) ) is calculated by multiplying the magnetic field (\( B \) ) by the perpendicular area (\( A \) ) through which the field passes:

• \[\Phi = B \cdot A \\]
• The unit of magnetic flux is Webers (Wb).

In the context of coils, as given in the exercise, the magnetic flux through each turn of the coil is concerned. Here, it's calculated using mutual inductance (\( M \) ) and the current in the coil:

• \[\Phi = \frac{MI}{N} \\]
• where \( I \) is the current and \( N \) is the number of turns in the coil.

Understanding magnetic flux helps in determining how efficiently a coil can induce an electromotive force when exposed to changing magnetic fields.

Faraday's Law of Induction

Faraday's Law of Induction explains how a changing magnetic flux can induce an electromotive force (emf) in a circuit. It is one of the foundational principles of electromagnetism. Essentially, it describes how the emf generated is related to the rate of change of magnetic flux through a circuit:

• \[ \mathcal{E} = - \frac{d\Phi}{dt} \]
• The negative sign signifies Lenz's law, indicating the direction of the induced emf opposes the change in flux.
• This principle is critical in understanding transformers, electric generators, and inductors.

In the exercise, Faraday's Law helps calculate the induced emf in a secondary coil when there is a change in current in a primary coil. This relationship is quantifiable due to the mutual inductance between the coils.

Induced EMF

An induced electromotive force (emf) is generated when there is a change in magnetic environment around a conductor, like the coils in our exercise. According to Faraday's Law, this induced emf (\( \mathcal{E} \) ) is directly dependent on the rate at which the magnetic flux changes. For two mutually coupled coils, the expression becomes:

• \[ \mathcal{E} = M \frac{dI}{dt} \]
• where \( M \) is the mutual inductance and \( \frac{dI}{dt} \) is the rate of change of current.

This concept is precisely how transformers work, allowing power transmission over long distances. In the given problem, it explains how a changing current in one coil results in an emf in the other coil, emphasizing the importance of mutual inductance. Understanding this helps grasp how energy transfer occurs in electromagnetic fields.