/*! 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 134 A coil having \(n\) turns and re... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 134

A coil having \(n\) turns and resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 R \Omega\). This combination is moved in time \(t\) seconds from a magnetic field \(W_{1}\) weber to \(W_{2}\) weber. The induced current in the circuit is (A) \(-\frac{\left(W_{2}-W_{1}\right)}{R n t}\) (B) \(-\frac{n\left(W_{2}-W_{1}\right)}{5 R t}\) (C) \(-\frac{\left(W_{2}-W_{1}\right)}{5 R n t}\) (D) \(-\frac{n\left(W_{2}-W_{1}\right)}{R t}\)

Short Answer

Expert verified
The induced current in the circuit is -\(\frac{n(W_{2}-W_{1})}{5Rt}\).

Step by step solution

01

Determine the change in magnetic flux

According to Faraday's law of electromagnetic induction, the electromotive force (EMF) induced in a loop is proportional to the rate of change of magnetic flux through the loop. The magnetic flux (桅) can be calculated by multiplying the magnetic field (W) by the area of the loop and the number of turns (n). Since the magnetic field changes from W1 to W2, we need to calculate the change in magnetic flux: 螖桅 = n (W2 - W1)
02

Calculate induced EMF using Faraday's law

Faraday's law states that the induced EMF (蔚) is equal to the negative rate of change of magnetic flux: 蔚 = -\(\frac{d(螖桅)}{dt}\) In this problem, the change in flux happens over time t, so we can simplify the expression as: 蔚 = -\(\frac{螖桅}{t}\) Now substituting the previously calculated change in magnetic flux: 蔚 = -\(\frac{n(W_{2}-W_{1})}{t}\)
03

Apply Ohm's Law to find induced current

Ohm's law states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is equal to the induced EMF, and the total resistance is the sum of the coil resistance and the galvanometer resistance: I = \(\frac{蔚}{R + 4R}\) Substituting the induced EMF and the total resistance: I = -\(\frac{n(W_{2}-W_{1})}{t(5R)}\) This matches option (B): -\(\frac{n(W_{2}-W_{1})}{5Rt}\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

A circular loop of radius \(0.3 \mathrm{~cm}\) lies parallel to a much bigger circular loop of radius \(20 \mathrm{~cm}\). The centre of the small loop is on the axis of the bigger loop. The distance between their centres is \(15 \mathrm{~cm}\). If a current of \(2.0 \mathrm{~A}\) flows through the smaller loop, then the flux linked with bigger loop is (A) \(6 \times 10^{-11}\) weber (B) \(3.3 \times 10^{-11}\) weber (C) \(6.6 \times 10^{-9}\) weber (D) \(9.1 \times 10^{-11}\) weber

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(I_{1}\) and \(I_{2}\), respectively. The magnitude of the magnetic induction at a point \(P\) at a distance \(a\) from the point \(O\) in a direction perpendicular to the plane \(A B C D\) is (A) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}+I_{2}\right)\) (B) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}-I_{2}\right)\) (C) \(\frac{\mu_{0}}{2 \pi a}\left(I_{1}^{2}+I_{2}^{2}\right)^{1 / 2}\) (D) \(\frac{\mu_{0}}{2 \pi a}\left(\frac{I_{1} I_{2}}{I_{1}+I_{2}}\right)\)

An iron rod of cross-sectional area 4 sq \(\mathrm{cm}\) is placed with its length parallel to a magnetic field of intensity \(1600 \mathrm{amp} / \mathrm{m}\). The flux through the rod is \(4 \times 10^{-4}\) weber. The permeability of the material of the rod is (In weber/amp-m). (A) \(0.625\) (B) \(6.25\) (C) \(0.625 \times 10^{-3}\) (D) None of these

A capacitance \(C\) is connected to a conducting rod of length \(\ell\) moving with a velocity \(v\) in a transverse magnetic field \(B\) then the charge developed in the capacitor is (A) Zero (B) \(B \ell v C\) (C) \(\frac{B l v C}{2}\) (D) \(\frac{B l v C}{3}\)

A magnetic field \(\vec{B}=\left(\frac{B_{0} y}{a}\right) \hat{k}\) is into the paper in the \(+z\) direction. \(B_{0}\) and \(a\) are positive constants. A square loop \(E F G H\) of side \(a\), mass \(m\), and resistance \(R\), in \(x-y\) plane, starts falling under the influence of gravity. Assume \(x\)-axis is horizontal and \(y\) is vertically downward. The expression for the speed of the loop \(v(t)\) is (A) \(\frac{m g}{B_{0}^{2} a^{2}}\left(1-e^{-\frac{B^{2} a^{2} t}{m R}}\right)\) (B) \(\frac{R m g}{B_{0}^{2} a^{2}}\left(1-e^{-\frac{B^{2} a^{2} t}{m R}}\right)\) (C) \(\frac{R m g}{B_{0}^{2} a^{2}}\left(e^{-\frac{B_{0}^{2} a^{2} t}{m R}}\right)\) (D) None of these
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Space Physics

Read Explanation

Magnetism

Read Explanation

Astrophysics

Read Explanation

Famous Physicists

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.