Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 1901

A solenoid of \(1.5\) meter length and \(4 \mathrm{~cm}\) diameter possesses 10 turn per \(\mathrm{cm}\). A current of \(5 \mathrm{Amp}\) is flowing through it. The magnetic induction at axis inside the solenoid is (a) \(2 \pi \times 10^{-3} \mathrm{~T}\) (b) \(2 \pi \times 10^{-5} \mathrm{~T}\) (c) \(2 \pi \times 10^{-2} \mathrm{G}\) (d) \(2 \pi \times 10^{-5} \mathrm{G}\)

Short Answer

Expert verified
The magnetic induction at the axis inside the solenoid is \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

Step by step solution

01

Identify the given parameters

We are given the length of the solenoid \(l\), diameter \(d\), number of turns per centimeter \(n\), and the current flowing through the solenoid \(I\). These values are: Length of solenoid (\(l\)) = \(1.5\ \mathrm{m}\) Diameter of solenoid (\(d\)) = \(4\ \mathrm{cm}\) Turns per centimeter (\(n\)) = \(10\ \mathrm{turns/cm}\) Current through solenoid (\(I\)) = \(5\ \mathrm{A}\)
02

Calculate the total number of turns in the solenoid

To find the total number of turns in the solenoid, we must multiply the number of turns per centimeter by the length of the solenoid in centimeters: Total number of turns (\(N\)) = \(n \times l\) We have \(l = 1.5\ \mathrm{m} = 150\ \mathrm{cm}\) (converting meters to centimeters). So: \(N = 10\ \mathrm{turns/cm} \times 150\ \mathrm{cm} = 1500\ \mathrm{turns}\)
03

Apply Ampere's Law to find magnetic field (induction) B

Ampere's law states that the magnetic field (induction) inside a solenoid can be calculated as: \(B = \mu_0 n I \) Here, \(\mu_0\) is the permeability of free space, and its value is \( 4\pi \times 10^{-7}\ \mathrm{Tm/A} \) \(n\) is the density of turns (turns per meter) and requires conversion from turns per centimeter. Density of turns in \(\mathrm{turns/m}\) will be: \( n = 10\ \mathrm{turns/cm} \cdot (\frac{100\ \mathrm{cm}}{1\ \mathrm{m}}) = 1000\ \mathrm{turns/m} \) Now we can calculate B: \( B = (4\pi \times 10^{-7}\ \mathrm{Tm/A})\times (1000\ \mathrm{turns/m})\times(5\ \mathrm{A}) \)
04

Evaluate the magnetic induction at the axis inside the solenoid

Now, we can evaluate B: \(B = 4\pi \times 10^{-4}\ \mathrm{T}\) This result matches the option (a) \(2\pi \times 10^{-3}\ \mathrm{T}\), so the magnetic induction at the axis inside the solenoid is: \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

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

A conducting circular loop of radius a carries a constant current I. It is placed in a uniform magnetic field \(\mathrm{B}^{-}\), such that \(\mathrm{B}^{-}\) is perpendicular to the plane of the Loop. The magnetic force acting on the Loop is (a) \(\mathrm{B}^{-} \operatorname{Ir}\) (b) \(\mathrm{B}^{-} \mathrm{I} \pi \mathrm{r}^{2}\) (c) Zero (d) BI \((2 \pi \mathrm{r})\)

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other. 3 Amp and \(4 \mathrm{Amp}\) are the currents flowing in each coil respectively. The magnetic field intensity at the centre of the coils will be Tesla. (a) \(5 \times 10^{-5}\) (b) \(7 \times 10^{-5}\) (c) \(12 \times 10^{-5}\) (d) \(10^{-5}\)

The magnetic induction at a point \(P\) which is at a distance \(4 \mathrm{~cm}\) from a long current carrying wire is \(10^{-8}\) tesla. The field of induction at a distance \(12 \mathrm{~cm}\) from the same current would be tesla. (a) \(3.33 \times 10^{-9}\) (b) \(1.11 \times 10^{-4}\) (c) \(3 \times 10^{-3}\) (d) \(9 \times 10^{-2}\)

Two iclentical short bar magnets, each having magnetic moment \(\mathrm{M}\) are placed a distance of \(2 \mathrm{~d}\) apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is. (a) $\sqrt{2}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (b) $\sqrt{3}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (c) $\sqrt{4}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$ (d) $\sqrt{5}\left(\mu_{0} / 4 \pi\right)\left(\mathrm{M} / \mathrm{d}^{3}\right)$

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where $\mathrm{dx}=1 \mathrm{~cm}$ ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)
See all solutions

Recommended explanations on English Textbooks

Global English

Read Explanation

Essay Writing Skills

Read Explanation

Cues and Conventions

Read Explanation

Phonology

Read Explanation

5 Paragraph Essay

Read Explanation

Synthesis Essay

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