/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A coil having an area \(A_{0}\) ... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 16

A coil having an area \(A_{0}\) is placed in a magnetic field which changes from \(B_{0}\) to \(4 B_{0}\) in time interval \(t\). The average EMF induced in the coil will be (A) \(\frac{3 A_{0} B_{0}}{t}\) (B) \(\frac{4 A_{0} B_{0}}{t}\) (C) \(\frac{3 B_{0}}{A_{0} t}\) (D) \(\frac{4 B_{0}}{A_{0} t}\)

Short Answer

Expert verified
The short answer is: The average EMF induced in the coil is \(\frac{3 A_{0} B_{0}}{t}\).

Step by step solution

01

Recall Faraday's Law of Electromagnetic Induction

Faraday's law states that the EMF induced in a coil is equal to the negative rate of change of the magnetic flux through the coil. Mathematically, it can be represented as: \(EMF = -\frac{\Delta \Phi}{\Delta t}\) where \(\Phi\) is the magnetic flux and \(\Delta t\) is the time interval. In this problem, we are asked to find the average EMF, so we will need to consider the average rate of change of the magnetic flux over the given time interval.
02

Define Magnetic Flux

Magnetic flux is the measure of the magnetic field that passes through a given area. For a coil with area A and magnetic field B, the magnetic flux can be calculated as: \(\Phi = A \times B\)
03

Calculate the Change in Magnetic Flux

In this problem, the magnetic field changes from B鈧� to 4B鈧�. To find the change in magnetic flux, we will subtract the initial magnetic flux from the final magnetic flux: \(\Delta\Phi = A_{0}(4B_{0}) - A_{0}(B_{0})\) After simplification, \(\Delta \Phi = 3A_{0}B_{0}\)
04

Calculate the Average EMF

Now, we can plug the change in magnetic flux and the time interval t into the Faraday's law equation: \(Average \,EMF = -\frac{\Delta\Phi}{\Delta t} = -\frac{3A_{0}B_{0}}{t}\) Since the negative sign indicates the direction of the induced EMF, which is not relevant to this exercise, we can remove the negative sign to find the magnitude of the average EMF: \(Average \,EMF = \frac{3A_{0}B_{0}}{t}\) Comparing with the provided options, the correct answer is: (A) \(\frac{3 A_{0} B_{0}}{t}\)

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Most popular questions from this chapter

A conducting square loop of side \(L\) and resistance \(R\) moves in its plane with a uniform velocity \(v\) perpendicular to one of its sides. A magnetic field \(B\), constant in space and time, pointing perpendicular and into the plane of the loop exists everywhere as shown in Fig. 16.38. The current induced in the loop is (A) \(B L v / R\) clockwise (B) \(B L v / R\) anti-clockwise (C) \(2 B L v / R\) anti-clockwise (D) Zero

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area \(A=10 \mathrm{~cm}^{2}\) and length \(=20 \mathrm{~cm} .\) If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is \(\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\right) \quad[\mathbf (A) \)2.4 \pi \times 10^{-5} \mathrm{H}\( (B) \)4.8 \pi \times 10^{-4} \mathrm{H}\( (C) \)4.8 \pi \times 10^{-5} \mathrm{H}\( (D) \)2.4 \pi \times 10^{-4} \mathrm{H}$

The self-inductance of the motor of an electric fan is \(10 \mathrm{H}\). In order to impart maximum power at \(50 \mathrm{~Hz}\), it should be connected to a capacitance of (A) \(8 \mu \mathrm{F}\) (B) \(4 \mu \mathrm{F}\) (C) \(2 \mu \mathrm{F}\) (D) \(1 \mu \mathrm{F}\)

A conducting circular loop is placed in a uniform magnetic field of induction \(B\) tesla with its plane normal to the field. Now the radius of the loop starts shrinking at the rate \((d r / d t)\). Then the induced EMF at the instant when the radius is \(r\) will be (A) \(\pi r B\left(\frac{d r}{d t}\right)\) (B) \(2 \pi r B\left(\frac{d r}{d t}\right)\) (C) \(\pi r^{2}\left(\frac{d B}{d t}\right)\) (D) \(B \frac{\pi r^{2}}{2} \frac{d r}{d t}\)

A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V}\). The current reaches half of its steady state value in (A) \(0.1 \mathrm{~s}\) (B) \(0.05 \mathrm{~s}\) (C) \(0.3 \mathrm{~s}\) (D) \(0.15 \mathrm{~s}\)
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