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Chapter 9: Problem 8

A perfectly conducting filament is formed into a circular ring of radius \(a\). At one point, a resistance \(R\) is inserted into the circuit, and at another a battery of voltage \(V_{0}\) is inserted. Assume that the loop current itself produces negligible magnetic field. ( \(a\) ) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; \((b)\) repeat part \(a\), assuming the battery is removed, the ring is closed again, and a linearly increasing \(\mathbf{B}\) field is applied in a direction normal to the loop surface.

Short Answer

Expert verified
Answer: With the battery, there is no induced EMF in the loop. Without the battery and with the increasing magnetic field, the induced EMF in the loop is -πa²k, where k is a constant representing the rate of increase of the magnetic field.

Step by step solution

01

Write down Faraday's Law

Faraday's Law states that the induced EMF (\(ε\)) in a loop is equal to the negative time rate of change of the magnetic flux (\(Φ\)) through the loop. Mathematically, this is: \(ε=-\dfrac{dΦ}{dt}\)
02

Analyze the circuit with the battery

In this scenario, the battery with voltage \(V_0\) is inserted. According to Kirchhoff's Voltage Law, the sum of the potential differences (voltages) around the loop is equal to zero. Considering the battery voltage and the voltage drop across the resistor (Ohm's Law), we can write: \(V_0 - IR - ε = 0\) where \(I\) is the current flowing through the resistor R.
03

Compute the EMF with the battery

Now, we need to find the induced EMF (\(ε\)) using Faraday's Law. Since the magnetic field is negligible, the rate of change of the magnetic field will be zero. According to Faraday's Law: \(ε = -\dfrac{dΦ}{dt} = 0\) Therefore, there is no induced EMF in the loop with the battery. This implies that: \(V_0 - IR = 0\) Without the battery:
04

Write the equation for the increasing magnetic field

In this case, it is given that a linearly increasing magnetic field (\(B(t)\)) is applied in a direction normal to the loop surface. We will assume the magnetic field to be in the form: \(B(t) = kt\) where \(k\) is a constant.
05

Calculate the magnetic flux

Now, we need to find the magnetic flux through the loop due to this magnetic field. The magnetic flux is given by: \(Φ = BAcos(θ)\) Since the magnetic field is perpendicular to the loop, we have \(θ = 0\), and the cosine term will become 1. The area of the circular loop is \(A = πa^2\). Therefore, the magnetic flux becomes: \(Φ(t) = (kt)(πa^2)\)
06

Calculate the induced EMF from the changing magnetic field

Applying Faraday's Law, we have: \(ε = -\dfrac{dΦ}{dt}\) To find the time derivative of the magnetic flux, we have: \(\dfrac{dΦ(t)}{dt} = πa^2\dfrac{d(kt)}{dt} = πa^2k\) Substituting this into Faraday's Law: \(ε = -\dfrac{dΦ}{dt} = -πa^2k\) This is the induced EMF in the loop without the battery and with the increasing magnetic field.

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Most popular questions from this chapter

Let the internal dimensions of a coaxial capacitor be \(a=1.2 \mathrm{~cm}, b=4 \mathrm{~cm}\), and \(l=40 \mathrm{~cm}\). The homogeneous material inside the capacitor has the parameters \(\epsilon=10^{-11} \mathrm{~F} / \mathrm{m}, \mu=10^{-5} \mathrm{H} / \mathrm{m}\), and \(\sigma=10^{-5} \mathrm{~S} / \mathrm{m}\). If the electric field intensity is \(\mathbf{E}=\left(10^{6} / \rho\right) \cos 10^{5} t \mathbf{a}_{\rho} \mathrm{V} / \mathrm{m}\), find \((a) \mathbf{J} ;(b)\) the total conduction current \(I_{c}\) through the capacitor; \((c)\) the total displacement current \(I_{d}\) through the capacitor; \((d)\) the ratio of the amplitude of \(I_{d}\) to that of \(I_{c}\), the quality factor of the capacitor.

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