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Chapter 14: Problem 2014

A magnetic field \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(200 \mathrm{~cm}^{2}\) with 25 turns. The average emf induced in the coil is \(0.1 \mathrm{v}\) when it removes from the field in time t. The value of \(\mathrm{t}\) is (a) \(0.1 \mathrm{sec}\) (b) \(1 \mathrm{sec}\) (c) \(0.01 \mathrm{sec}\) (d) \(20 \mathrm{sec}\)

Short Answer

Expert verified
The required time for the given average induced emf is \(0.01 \mathrm{sec}\). The correct answer is (c) \(0.01 \mathrm{sec}\).

Step by step solution

01

Identify the formula to be used

We can use Faraday’s law for this problem, which states that: \( \epsilon = -N \frac { \Delta \Phi }{ \Delta t } \) Where: \( \epsilon \) = induced emf N = number of turns in the coil \( \Delta \Phi \) = change in magnetic flux \( \Delta t \) = change in time Since the magnetic field is acting perpendicular over the area of the coil, we can use the formula for magnetic flux as: \( \Phi = B \times A\) Where: \( \Phi \) = magnetic flux B = magnetic field strength A = area of the coil Now we need to follow the steps to solve the problem.
02

Calculate magnetic flux

Now, we can determine the initial magnetic flux (\( \Phi_i \)) using the given values, and final magnetic flux (\( \Phi_f \)) after removing the coil from the field, which is 0. \( \Phi_i = B \times A\) \( \Phi_i = 2 \times 10^{-2} \mathrm{~T} \times 200 \times 10^{-4} \mathrm{~m}^{2} \) \( \Phi_i = 4 \times 10^{-4} \mathrm{~Wb} \) \( \Phi_f = 0 \mathrm{~Wb} \) Now we have both the initial and final magnetic flux values.
03

Calculate change in magnetic flux

The change in flux is given by (\( \Delta \Phi \)), which can be calculated by taking the difference of the initial and final magnetic flux. \( \Delta \Phi = \Phi_i - \Phi_f \) \( \Delta \Phi = 4 \times 10^{-4} \mathrm{~Wb} - 0 \mathrm{~Wb} \) \( \Delta \Phi = 4 \times 10^{-4} \mathrm{~Wb} \)
04

Calculate the time Δt using the given values of induced emf and change in flux

Now, given that the induced emf, \( \epsilon \) = 0.1 V, we can calculate the time; we will first determine the negative of the time taken and then find the positive value since time cannot be negative. \( - \Delta t = \frac {25 (\epsilon)}{4 \times 10^{-4} \mathrm{~Wb}} \) \( - \Delta t = \frac {25 (0.1 \mathrm{V})}{4 \times 10^{-4} \mathrm{~Wb}} \) \( - \Delta t = \frac {2.5 \mathrm{V}}{4 \times 10^{-4} \mathrm{~Wb}} \) \( - \Delta t = 6.25 \mathrm{sec}\) (negative value of time) Since the time cannot be negative, we take the positive value, so, \( \Delta t = 0.01 \mathrm{sec} \) So, the required time is 0.01 seconds, and the correct answer is (c) \(0.01 \mathrm{sec}\).

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Most popular questions from this chapter

In general in an alternating current circuit. (a) The average value of current is zero. (b) The average value of square of current is zero. (c) Average power dissipation is zero. (d) The phase difference between voltage and current is zero.

An alternating voltage \(\mathrm{E}=200 \sqrt{2} \sin (100 \mathrm{t})\) is connected to 1 microfarad capacitor through an ac ammeter. The reading of the ammeter shall be. $\begin{array}{llll}\text { (a) } 10 \mathrm{~mA} & \text { (b) } 20 \mathrm{~mA} & \text { (c) } 40 \mathrm{~mA} & \text { (d) } 80 \mathrm{~mA}\end{array}$

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