Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 2059

Two coils have a mutual inductance \(0.005 \mathrm{H}\). The current changes in a coil according to equation \(\mathrm{I}=\mathrm{I}_{0}\) sin ot. Where \(I_{0}=10 \mathrm{~A} \& \omega=100 \pi \mathrm{rad} \mathrm{s}^{-1}\). The maximum value of emf in second coil is ............. (a) \(2 \pi\) (b) \(5 \pi\) (c) \(\pi\) (d) \(4 \pi\)

Short Answer

Expert verified
The maximum value of emf in the second coil is \(5\pi\).

Step by step solution

01

We are given the mutual inductance M = 0.005 H, the maximum current \(I_0 = 10 A\), the angular frequency \(\omega = 100\pi rad/s\), and the equation of the current change in the first coil: \(I = I_0\sin(\omega t)\). #Step 2: Apply Faraday's law of electromagnetic induction#

According to Faraday's law of electromagnetic induction, the induced emf in the second coil is given by \(-\frac{d\Phi}{dt}\), where \(\Phi\) is the mutual magnetic flux. The mutual magnetic flux is the product of mutual inductance (M) and current (I) in the first coil, i.e., \(\Phi = MI = M I_0\sin(\omega t)\). #Step 3: Compute the time derivative of the mutual magnetic flux#
02

To find the induced emf, first compute the time derivative of the magnetic flux: \(\frac{d\Phi}{dt} = M I_0 \omega\cos(\omega t)\). #Step 4: Find the maximum value of induced emf in the second coil#

The maximum value of the induced emf can be found when the cosine term in the time derivative of the magnetic flux is at its maximum (\(\cos(\omega t) = 1\)). Therefore, the maximum induced emf is: \(\frac{d\Phi}{dt_{max}} = MI_0\omega\). #Step 5: Substitute the given values and calculate the maximum induced emf#
03

Now, substitute the given values of M, \(I_0\), and \(\omega\) in the equation: \(Max\:emf = 0.005\cdot 10\cdot 100\pi\). Calculate the result: \(Max\:emf = 5\pi\). #Step 6: Choose the correct option#

Based on our calculations, the maximum value of emf in the second coil is \(5\pi\). Therefore, the correct option is (b) \(5 \pi\).

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

The time constant of a LR circuit is \(80 \mathrm{~ms}\), The circuit is connected at \(t=0\) and the steady state current is found to be $4 \mathrm{~A}\(. Find the current at \)20 \mathrm{~ms}$. (a) \(0.98 \mathrm{~A}\) (b) \(1 \mathrm{~A}\) (c) \(0.44 \mathrm{~A}\) (d) \(0.88 \mathrm{~A}\)

In a LCR circuit capacitance is changed from \(\mathrm{C}\) to \(2 \mathrm{C}\). For the resonant frequency to remain unchanged, the inductance should be change from \(L\) to (a) \(4 \mathrm{~L}\) (b) \(2 \mathrm{~L}\) (c) \(L / 2\) (d) \(\mathrm{L} / 4\)

An inductor-resistor-battery circuit is switched on at \(\mathrm{t}=0 .\) If the emf of battery is \(\varepsilon\) find the charge passes through the battery in one time constant \(\tau\). (a) \(\left[\left\\{i_{\max } \tau\right\\} / \mathrm{e}\right]\) (b) \((\tau-1) \mathrm{e} \mathrm{i}_{\max }\) (c) \(\left[\left\\{i_{\max }\right\\} / \mathrm{e}\right]\) (d) \(\tau \mathrm{i}_{\max }\)

A conducting rod \(P Q\) of length \(4 \ell\) is rotated about a point \(O\) in a uniform magnetic field \(\mathrm{B} \rightarrow \mathrm{PO}=\ell\) Then (a) $\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{P}}=-\left[\left\\{\mathrm{B} \omega \ell^{2}\right\\} /\right.$ \(\\{2\\}] \quad\) (b) $\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{O}}=(5 / 2) \mathrm{B} \operatorname{co} \ell^{2}$ (c) $\mathrm{V}_{Q}-\mathrm{V}_{\mathrm{O}}=(9 / 2) \mathrm{B} \omega \ell^{2}$ (d) $\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}=4 \mathrm{~B} \omega \ell^{2}$

An alternating current of \(\mathrm{rms}\) value \(10 \mathrm{~A}\) is passed through a \(12 \Omega\) resistance. The maximum potential difference across the resistor is, (a) \(20 \mathrm{~V}\) (b) \(90 \mathrm{~V}\) (c) \(169.68 \mathrm{~V}\) (d) None of these
See all solutions

Recommended explanations on English Textbooks

Essay Prompts

Read Explanation

History of English Language

Read Explanation

Pragmatics

Read Explanation

Sociolinguistics

Read Explanation

Creative Story

Read Explanation

Graphology

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