Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 1896

Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of the magnetic moment is (a) \(2: \pi\) (b) \(\pi: 2\) (c) \(\pi: 4\) (d) \(4: \pi\)

Short Answer

Expert verified
The ratio of the magnetic moment of the square to the circle is \(\pi: 2\), which corresponds to option (b).

Step by step solution

01

Calculate the magnetic moment of the square wire

Let's denote the length of the wire as L. Since the wire is shaped into a square, each side of the square will have a length of L/4. The area A_s of the square can be calculated as: \[A_s = (\frac{L}{4})^2\] Now, let's denote the current passing through the wire as I. The magnetic moment M_s of the square can be calculated as: \[M_s = I \cdot A_s\] By substituting the area of the square, we get: \[M_s = I \cdot (\frac{L}{4})^2\]
02

Calculate the magnetic moment of the circle wire

To find the magnetic moment of the circle wire, we first need to determine its radius. Since the circumference of the circle is equal to the length of the wire, we have: \[2\pi r = L\] Solving for the radius r, we get: \[r = \frac{L}{2\pi}\] Now, we can calculate the area A_c of the circle as: \[A_c = \pi r^2\] Substituting the radius, we get: \[A_c = \pi (\frac{L}{2\pi})^2\] The magnetic moment M_c of the circle can be calculated as: \[M_c = I \cdot A_c\] By substituting the area of the circle, we get: \[M_c = I \cdot \pi (\frac{L}{2\pi})^2\]
03

Calculate the ratio of the magnetic moments

We need to find the ratio of the magnetic moments of the square and the circle: \[\frac{M_s}{M_c} = \frac{I \cdot (\frac{L}{4})^2}{I \cdot \pi (\frac{L}{2\pi})^2}\] We can see that the current I and the length L will cancel out, so the ratio of the magnetic moments simplifies to: \[\frac{M_s}{M_c} = \frac{(\frac{1}{4})^2}{\pi (\frac{1}{2\pi})^2}\] After calculating the ratio, we arrive at the following: \[\frac{M_s}{M_c} = \frac{1}{16} \cdot \frac{4\pi^2}{\pi}\] This further simplifies to: \[\frac{M_s}{M_c} = \frac{\pi}{2}\] So the ratio of the magnetic moment of the square to the circle is \(\pi: 2\), which corresponds to option (b).

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 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) \(10^{-5}\)

A proton is projected with a speed of $2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\( at an angle of \)60^{\circ}\( to the \)\mathrm{X}$ -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $\pi \times 10^{-7} \mathrm{sec}$ (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (c) A helix of \(r=0.1 \mathrm{~m}\) and time period $2 \pi \times 10^{-7} \mathrm{sec}$ (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period $4 \pi \times 10^{-7} \mathrm{sec}$

\(\mathrm{A}\) bar magnet of length \(10 \mathrm{~cm}\) and having the pole strength equal \(10^{3} \mathrm{Am}\) to is kept in a magnetic field having magnetic induction (B) equal to \(4 \pi \times 10^{3}\) tesla. It makes an angle of \(30^{\circ}\) with the direction of magnetic induction. The value of the torque acting on the magnet is Joule. (a) \(2 \pi \times 10^{-7}\) (b) \(2 \pi \times 10^{5}\) (c) \(0.5\) (d) \(0.5 \times 10^{2}\)

A Galvanometer coil has a resistance of \(15 \Omega\) and gives full scale deflection for a current of \(4 \mathrm{~mA}\). To convert it to an ammeter of range 0 to \(6 \mathrm{Amp}\) (a) \(10 \mathrm{~m} \Omega\) resistance is to be connected in parallel to the galvanometer. (b) \(10 \mathrm{~m} \Omega\) resistance is to be connected in series with the galvanometer. (c) \(0.1 \Omega\) resistance is to be connected in parallel to the galvanometer. (d) \(0.1 \Omega\) resistance is to be connected in series with the galvanometer.

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25
See all solutions

Recommended explanations on English Textbooks

Single Paragraph Essay

Read Explanation

Language Acquisition

Read Explanation

Language and Social Groups

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Synthesis Essay

Read Explanation

International English

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