Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 1869

The strength of the magnetic field at a point \(\mathrm{y}\) near a long straight current carrying wire is \(\mathrm{B}\). The field at a distance \(\mathrm{y} / 2\) will be (a) B/2 (b) B \(/ 4\) (c) \(2 \mathrm{~B}\) (d) \(4 \mathrm{~B}\)

Short Answer

Expert verified
The magnetic field strength at a distance \(y/2\) from the long straight current-carrying wire is twice the magnetic field strength at distance \(y\). Therefore, the correct answer is option (c): \(B' = 2B\).

Step by step solution

01

Identify the formula for magnetic field strength

The formula for the magnetic field strength (B) near a long straight current-carrying wire at a distance (r) is given by the following relation from the Biot-Savart law: \[B = \frac{\mu_0 I}{2\pi r}\] where \(\mu_0\) is the permeability of free space, \(I\) is the current through the wire, and \(r\) is the distance from the wire to the point where we wish to calculate the magnetic field strength.
02

Calculate the magnetic field strength at distance 'y'

We are asked to find the magnetic field strength at distance \(y\) and \(y/2\). Let's first find the magnetic field strength at distance \(y\), which is given as \(B\). We can derive the following equation from the formula above: \[B = \frac{\mu_0 I}{2\pi y}\]
03

Calculate the magnetic field strength at distance 'y/2'

Now let's calculate the magnetic field strength at distance \(y/2\), which we can denote as \(B'\). We'll substitute the distance as \(y/2\) in the formula for magnetic field strength: \[B' = \frac{\mu_0 I}{2\pi(\frac{y}{2})}\]
04

Find the relationship between B and B'

We need to find the relationship between \(B\) and \(B'\). Divide the equation of \(B'\) by the equation of \(B\): \[\frac{B'}{B} = \frac{\frac{\mu_0 I}{2\pi(\frac{y}{2})}}{\frac{\mu_0 I}{2\pi y}}\] We can see that \(\mu_0 I\) term cancels out in the numerator and denominator: \[\frac{B'}{B} = \frac{1}{\frac{1}{2}}\]
05

Solve the relationship between B and B'

Now, we can simplify the final equation: \[\frac{B'}{B} = 2\] So, we see that the magnetic field strength at distance \(y/2\) is twice the magnetic field strength at distance \(y\), which is option (c): \(B' = 2B\).

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 thin long parallel wires separated by a distance \(\mathrm{y}\) are carrying a current I Amp each. The magnitude of the force per unit length exerted by one wire on other is (a) \(\left[\left(\mu_{0} I^{2}\right) / y^{2}\right]\) (b) \(\left[\left(\mu_{o} I^{2}\right) /(2 \pi \mathrm{y})\right]\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi)\right](1 / y)\) (d) $\left[\left(\mu_{0}\right) /(2 \pi)\right]\left(1 / \mathrm{y}^{2}\right)$

The forces existing between two parallel current carrying conductors is \(\mathrm{F}\). If the current in each conductor is doubled, then the value of force will be (a) \(2 \mathrm{~F}\) (b) \(4 \mathrm{~F}\) (c) \(5 \mathrm{~F}\) (d) \((\mathrm{F} / 2)\)

The mag. field due to a current carrying circular Loop of radius $3 \mathrm{~cm}\( at a point on the axis at a distance of \)4 \mathrm{~cm}$ from the centre is \(54 \mu \mathrm{T}\) what will be its value at the centre of the LOOP. (a) \(250 \mu \mathrm{T}\) (b) \(150 \mu \mathrm{T}\) (c) \(125 \mu \mathrm{T}\) (d) \(75 \mu \mathrm{T}\)

In the case of bar magnet, lines of magnetic induction (a) Start from the North pole and end at the South pole (b) Run continuously through the bar and outside (c) Emerge in circular paths from the middle of the bar (d) Are produced only at the North pole like rays of light from a bulb.

A Galvanometer of resistance \(15 \Omega\) is connected to a battery of 3 volt along with a resistance of \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) \(6050 \Omega\) (b) \(4450 \Omega\) (c) \(5050 \Omega\) (d) \(5550 \Omega\)
See all solutions

Recommended explanations on English Textbooks

Argumentative Essay

Read Explanation

Creative Story

Read Explanation

Summary Text

Read Explanation

Essay Writing Skills

Read Explanation

Lexis and Semantics

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