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A 1000-turn coil of wire 2.0 cm in diameter is in a magnetic field that drops from \(0.10 \mathrm{T}\) to \(0 \mathrm{T}\) in \(10 \mathrm{ms}\). The axis of the \(\mathrm{coil}\) is parallel to the field. What is the emf of the coil?

Short Answer

Expert verified
Given the required inputs, substituting N, A, ΔB, and Δt into Faraday's law formula to derive the value of the induced electromotive force (emf), ε.

Step by step solution

01

Calculate the coil's area

Use the diameter of the coil to calculate the coil's area using the formula A = πr² = π(D/2)², where D is the diameter. Given D = 2.0 cm, we convert to meters since standard SI unit for magnetic field B is T(m tesla) which requires meters for length.
02

Find the change in the magnetic field

The change in the magnetic field, ΔB, is determined from the difference between the final magnetic field( Bf) and initial magnetic field (Bi). In our case, ΔB = Bf - Bi = 0T - 0.10T = -0.10T assuming away from the coil is positive, therefore, the negative sign indicates the field is increasing towards the coil.
03

Faraday's Law

Using Faraday's law of electromagnetic induction, we calculate the induced emf. The law is given by ε = -N * (ΔΦ/Δt) = -N * (A * ΔB/Δt), where Φ is magnetic flux (equal to A * B). Substiute the values: N = 1000 turns, Δt = 0.01 seconds (from 10 ms), A from Step 1 and ΔB from Step 2. Ensure the units are consistent before calculating.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Flux
Magnetic flux is a crucial concept when discussing electromagnetic induction. It provides a measure of the total magnetic field passing through a given area, like a coil. In mathematical terms, magnetic flux (Φ) is defined as the product of the magnetic field (B) and the perpendicular area (A) it penetrates. The formula is \[ \Phi = B \cdot A \cdot \cos(\theta) \]where:
  • \(\Phi\) is the magnetic flux,
  • \(B\) is the magnetic field (in teslas, T),
  • \(A\) is the area (in square meters, m²),
  • \(\theta\) is the angle between the field lines and the normal to the surface.
For our exercise, the coil is parallel to the field, making the angle \(\theta = 0\), which simplifies the equation to \(\Phi = B \cdot A\). As the magnetic field changes, so does the magnetic flux, leading to the induction of an electromotive force (EMF).
Faraday's Law
Faraday's Law is the fundamental principle behind electromagnetic induction. It describes how a changing magnetic flux induces an electromotive force (EMF) in a closed circuit. The law is represented by the equation\[ \varepsilon = -N \cdot \frac{\Delta \Phi}{\Delta t} \]where:
  • \(\varepsilon\) is the induced EMF,
  • \(N\) is the number of turns in the coil,
  • \(\Delta \Phi\) is the change in magnetic flux,
  • \(\Delta t\) is the time over which the change occurs.
The negative sign in the formula indicates Lenz's Law, showing that the induced EMF opposes the change in flux. In our exercise, you observe this as the magnetic field drops from \(0.10 \mathrm{T}\) to \(0 \mathrm{T}\) over \(10 \mathrm{ms}\). This rapid change induces an EMF which can be calculated using the above formulation.
Induced EMF
Induced EMF is a central concept that emerges from a changing magnetic environment, governed by Faraday's Law. It is the voltage generated across the coil as a result of changing magnetic flux. In our scenario, the EMF is computed as the magnetic field goes to zero. Using Faraday's Law, \(\varepsilon = -N \cdot \frac{\Delta (B \cdot A)}{\Delta t}\), we deduce that the changing magnetic field and area of the coil determines the induced EMF.Once you calculate the coil's area and ascertain the magnetic flux change, this formula enables the computation of the induced EMF efficiently. This process highlights the dynamic relationship between a magnetic environment and its resulting electrical effects.
Coil Area Calculation
Calculating the coil area is a straightforward but essential step in determining the induced EMF. The area of a coil is vital for computing magnetic flux and understanding the impact of the magnetic field on the coil.The area of a circular coil is found using the formula:\[ A = \pi r^2 = \pi \left( \frac{D}{2} \right)^2 \]where:
  • \(A\) is the area of the coil,
  • \(r\) is the radius of the coil,
  • \(D\) is the diameter of the coil.
In this exercise, with a diameter of \(2.0 \mathrm{cm}\), convert it to meters (0.02 m) for calculations, since magnetic field calculations require SI units. The calculated area plays a crucial role in determining the total magnetic flux, and ultimately, the induced EMF in the coil.

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Most popular questions from this chapter

A 40-turn, 4.0-cm-diameter coil with \(R=0.40 \Omega\) surrounds a 3.0-cm- diameter solenoid. The solenoid is 20 cm long and has 200 turns. The \(60 \mathrm{Hz}\) current through the solenoid is \(I=\) \(I_{0} \sin (2 \pi f t) .\) What is \(I_{0}\) if the maximum induced current in the coil is \(0.20 \mathrm{A} ?\)

The maximum charge on the capacitor in an oscillating \(L C\) circuit is \(Q_{0}\). What is the capacitor charge, in terms of \(Q_{0}\), when the energy in the capacitor's electric field equals the energy in the inductor's magnetic field?

The current through inductance \(L\) is given by \(I=I_{0} \sin \omega t .\) a. Find an expression for the potential difference \(\Delta V_{\mathrm{L}}\) across the inductor. b. The maximum voltage across the inductor is \(0.20 \mathrm{V}\) when \(L=50 \mu \mathrm{H}\) and \(f=500 \mathrm{kHz} .\) What is \(I_{0} ?\)

An \(L C\) circuit is built with a \(20 \mathrm{mH}\) inductor and an \(8.0 \mu \mathrm{F}\) capacitor. The current has its maximum value of 0.50 A at \(t=0\) s. a. How long is it until the capacitor is fully charged? b. What is the voltage across the capacitor at that time?

You need to make a \(100 \mu \mathrm{H}\) inductor on a cylinder that is \(5.0 \mathrm{cm}\) long and \(1.0 \mathrm{cm}\) in diameter. You plan to wrap four layers of wire around the cylinder. What diameter wire should you use if the coils are tightly wound with no space between them? The wire diameter is small enough you don't need to consider the change in the coil's diameter for the outer layers.

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