Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 2019

A coil having n turns \(\&\) resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 \mathrm{R} \Omega\). This combination is moved from a magnetic field \(\mathrm{W}_{1} \mathrm{~Wb}\) to $\mathrm{W}_{2} \mathrm{~Wb}\( in \)\mathrm{t}$ second. The induced current in the circuit is.... (a) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rnt}\\}\right]$ (b) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rt}\\}\right]$ (c) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rnt}\\}\right]$ (d) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rt}\\}\right]$

Short Answer

Expert verified
The short answer is: \(I = -n\left[\frac{W_2-W_1}{Rt}\right]\).

Step by step solution

01

Calculate the change in magnetic flux

First, find the change in magnetic flux. The magnetic flux can be calculated as the product of the number of turns \(n\) and the change in magnetic field \(\Delta W = W_2 - W_1\): \[\Delta \Phi = n\Delta W\]
02

Apply Faraday's law of electromagnetic induction

According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) in the circuit is equal to the negative rate of change of magnetic flux: \[EMF = -\frac{d\Phi}{dt}\] Since the change in magnetic flux happens in \(t\) seconds, we replace \(d\Phi\) with \(\Delta \Phi\) and \(dt\) with \(t\): \[EMF = -\frac{\Delta \Phi}{t}\]
03

Substitute the values from Step 1 into the equation from Step 2

Substitute the values from Step 1 into the equation from Step 2: \[EMF = -\frac{n\Delta W}{t}\]
04

Apply Ohm's law to find the induced current

According to Ohm's law, the induced current in the circuit is equal to the induced EMF divided by the total resistance in the circuit: \[I = \frac{EMF}{R_t}\] The total resistance in the circuit is the sum of the resistance of the coil and the galvanometer: \[R_t = R + 4R = 5R\]
05

Substitute the values from Steps 3 and 4 to find the induced current

Substitute the values from Steps 3 and 4 into the equation for the induced current: \[I = \frac{-n\Delta W}{5Rt}\] We can rewrite \(\Delta W\) as \(W_2 - W_1\): \[I = -\frac{n(W_2-W_1)}{5Rt}\] This expression matches option (d) among the given choices. Thus, the induced current in the circuit is: \[I = -n\left[\frac{W_2-W_1}{Rt}\right]\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two co-axial solenoids are made by a pipe of cross sectional area $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...... (a) \(4.8 \pi \times 10^{-4} \mathrm{H}\) (b) \(4.8 \pi \times 10^{-5} \mathrm{H}\) (c) \(2.4 \pi \times 10^{-4} \mathrm{H}\) (d) \(4.8 \pi \times 10^{4} \mathrm{H}\)

A resistor and a capacitor are connected in series with an ac source. If the potential drop across the capacitor is \(5 \mathrm{~V}\) and that across resistor is \(12 \mathrm{~V}\), the applied voltage is, (a) \(13 \mathrm{~V}\) (b) \(17 \mathrm{~V}\) (c) \(5 \mathrm{~V}\) (d) \(12 \mathrm{~V}\)

The quality factor of LCR circuit having resistance \((\mathrm{R})\) and inductance ( \(\mathrm{L}\) ) at resonance frequency \((\infty)\) is given by (a) \((\mathrm{cL} / \mathrm{R})\) (b) \((\mathrm{R} / \mathrm{cL})\) (c) \((\mathrm{coL} / \mathrm{R})^{(1 / 2)}\) (d) \((0 / L)^{2}\)

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.

A coil having area \(2 \mathrm{~m}^{2}\) is placed in a magnetic field which changes from \(1 \mathrm{wb} / \mathrm{m}^{2}\) to $4 \mathrm{wb} / \mathrm{m}^{2}$ in an interval of 2 second. The emf induced in the coil of single turn is.... (a) \(4 \mathrm{v}\) (b) \(3 \mathrm{v}\) (c) \(1.5 \mathrm{v}\) (d) \(2 \mathrm{v}\)
See all solutions

Recommended explanations on English Textbooks

5 Paragraph Essay

Read Explanation

Argumentative Essay

Read Explanation

Essay Plan

Read Explanation

Prosody

Read Explanation

Listening and Speaking

Read Explanation

Lexis and Semantics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free