/*! 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 150 Two coaxial solenoids are made b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 150

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}$

Short Answer

Expert verified
The mutual inductance is \(2.4 \pi \times 10^{-4} \mathrm{H}\), so the correct answer is (D) \(2.4 \pi \times 10^{-4} \mathrm{H}\)

Step by step solution

01

- Gather all given values

The values given in the problem are \(\mu_{0} = 4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\), \(N1 = 300\) turns, \(N2 = 400\) turns, \(A = 10 \mathrm{~cm}^{2}\) (convert this to \(m^{2}\) for formula compatibility, thus \(A = 10 *10^{-4} \mathrm{m}^{2}\)), and \( l = 20 \mathrm{~cm}\) (also converted to m, so \(l = 20 * 10^{-2} \mathrm{m}\)).
02

- Apply the Mutual induction formula

The mutual induction formula is \(M = \frac{{\mu_{0} * N1 * N2 * A}}{l}\). Insert the given values to this formula: \(M = \frac{{4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1} * 300 * 400 * (10 *10^{-4} \mathrm{m}^{2})}}{20 * 10^{-2} \mathrm{m}}\).
03

- Calculate the result

Simplify the expression to get the result. After performing the multiplication and division, you get \(M = 2.4 \pi \times 10^{-4} \mathrm{H}\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Induction
Electromagnetic induction is a fascinating concept that allows us to create electricity from magnetic fields. This phenomenon occurs when a conductor, like a loop of wire, is placed in a magnetic field that is changing with time. Due to this change, an electromotive force (often abbreviated as EMF) is induced in the conductor, which can cause an electric current if the circuit is closed.

Think of it like a magic trick where movement of magnets (or a change in magnetic field) pulls electric currents out of seemingly thin air! Now, in the case of our solenoids from the exercise, when a current flows through one of the solenoids and changes (either increases or decreases), it induces a voltage in the adjacent solenoid, provided they are in close proximity and aligned along the same axis, which they are as 'coaxial'. This process of generating voltage due to a changing current in a nearby coil is a direct application of electromagnetic induction.
Coaxial Solenoids
Solenoids are just coils of wire, but when we say 'coaxial solenoids', we mean two solenoids that share the same central axis. Imagine two springs, one inside the other, along the same stretchy path. These coaxial solenoids are special because they interact with each other magnetically in a very efficient manner.

When an electric current passes through one solenoid, it generates a magnetic field. If another solenoid is coaxial to the first one, this magnetic field will pass through the second solenoid. Because these solenoids are aligned, the magnetic coupling between them is strong, which means a change in current in one solenoid will have a pronounced effect on the other, leading to a higher mutual inductance 鈥� which is precisely the concept we exploited in the exercise solution.
Inductance Calculation
The inductance calculation can seem a bit daunting, but it's essential for understanding how solenoids work together magnetically. Mutual inductance is the measure of how much a change in current in one solenoid will induce a voltage in another, which is expressed in units of Henrys (H). The mutual inductance, denoted by 'M', depends on a few factors: the permeability of free space \(\mu_0\), the number of turns in each solenoid \(N1\) and \(N2\), the cross-sectional area \(A\), and the length \(l\) of the solenoids.

The formula used in the exercise to calculate M is \(M = \frac{\mu_{0} * N1 * N2 * A}{l}\). This elegant equation contains all the critical variables. Our detailed walk-through in the solution involved converting centimeters to meters to ensure all units match the standard SI units for the formula to work correctly. Once values were plugged into the formula and the arithmetic complete, revealing the mutual inductance of the coaxial solenoids.

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 uniform conducting ring of mass \(\pi \mathrm{kg}\) and radius \(1 \mathrm{~m}\) is kept on smooth horizontal table. A uniform but time varying magnetic field \(B=\left(\hat{i}+t^{2} \hat{j}\right)\) Tesla is present in the region. (Where \(t\) is time in seconds). Resistance of ring is \(2 \Omega .\) Then, \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) Induced electric field (in volt/meter) at the circumference of ring at the instant ring starts toppling is (A) \(\frac{10}{\pi}\) (B) \(\frac{20}{\pi}\) (C) \(\frac{5}{\pi}\) (D) \(\frac{25}{\pi}\)

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) \(B \frac{\pi r^{2}}{2} \frac{d r}{d t}\)

Net force on a current carrying loop kept in uniform magnetic field is zero and the torque on the loop \(\vec{\tau}=\vec{M} \times \vec{B}\), where \(M\) and \(B\) are magnetic dipole moment and magnetic field intensity, respectively. If it is free to rotate, then it will rotates about an axis passing through its centre of mass and parallel to \(\vec{\tau}\). Potential energy of the loop is given by \(U=-\vec{M} \cdot \vec{B}\). Assume a current carrying ring with its centre at the origin and having moment of inertia \(2 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}\) about an axis passing through one of its diameter and magnetic moment \(\vec{M}=(3 \hat{i}-4 \hat{j}) \mathrm{Am}^{2}\). At time \(t=0\), a magnetic field \(\vec{B}=(4 \hat{i}-3 \hat{j}) T\) is switched on. Then Torque acting on the loop is (A) Zero (B) \(25 \hat{k} \mathrm{Nm}\) (C) \(16 \hat{k} \mathrm{Nm}\) (D) \(10 \hat{k} \mathrm{Nm}\)

A uniform conducting ring of mass \(\pi \mathrm{kg}\) and radius \(1 \mathrm{~m}\) is kept on smooth horizontal table. A uniform but time varying magnetic field \(B=\left(\hat{i}+t^{2} \hat{j}\right)\) Tesla is present in the region. (Where \(t\) is time in seconds). Resistance of ring is \(2 \Omega .\) Then, \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) Net magnetic force (in Newton) on conducting ring as function of time is (A) \(2 \pi^{2} t\) (B) \(2 \pi^{2} t^{2}\) (C) \(2 \pi^{2} t^{3}\) (D) Zero

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) \(\frac{\mu_{0}}{2 \pi a}\left(\frac{I_{1} I_{2}}{I_{1}+I_{2}}\right)\)
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Electricity

Read Explanation

Physical Quantities And Units

Read Explanation

Fundamentals of Physics

Read Explanation

Work Energy and Power

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.