/*! 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 57 Current \(i\) is carried in a wi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 57

Current \(i\) is carried in a wire of length \(L .\) If the wire is turned into a circular coil, the maximum magnitude of torque in a given magnetic field \(B\) will be (A) \(\frac{L^{2} i B}{2}\) (B) \(\frac{L^{2} i B}{\pi}\) (C) \(\frac{L^{2} i B}{4 \pi}\) (D) \(\frac{L i^{2} B}{4 \pi}\)

Short Answer

Expert verified
The short answer is: The maximum magnitude of torque in a given magnetic field \(B\) when the wire is turned into a circular coil is (C) \(\frac{L^{2} i B}{4 \pi}\).

Step by step solution

01

Calculate the circumference of the circular coil

As the wire is turned into a circular coil, the length of the wire will be equal to the circumference of the coil. The circumference of a circle is given by: \(C = 2 \pi r\), where r is the radius of the circle. Therefore, we can write: \(L = 2 \pi r\) Step 2: Calculate the area of the circle
02

Calculate the area of the circular coil

The area of a circle can be calculated using the formula: \(A = \pi r^2\). We will substitute the expression for the radius r from step 1 into this formula: \(A = \pi (\frac{L}{2\pi})^2 = \frac{L^2}{4\pi}\) Step 3: Calculate the magnetic moment
03

Calculate the magnetic moment of the circular coil

The magnetic moment of a coil carrying current (i) and having area (A) is given by: \(\mu = iA\). Using the area A found in step 2, we get: \(\mu = i\cdot\frac{L^2}{4\pi}\) Step 4: Calculate the maximum torque
04

Calculate the maximum torque on the circular coil

The torque experienced by a coil in a magnetic field is: \(\tau = \mu B\sin{\theta}\). The maximum torque is experienced when the angle between the magnetic moment and magnetic field is 90 degrees (i.e., sin(90) = 1). Therefore, the maximum torque is given by: \(\tau_{max} = \mu B = i \cdot \frac{L^2}{4\pi} B\) The answer is: (C) \(\frac{L^{2} i B}{4 \pi}\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque needed to maintain the needle in this position will be (A) \(\sqrt{3} W\) (B) \(W\) (C) \(\frac{\sqrt{3}}{2} W\) (D) \(2 W\)

An equilateral triangular current loop \(P Q R\) carries a current \(I\) ampere. Length of each side is \(l\) metre. A uniform magnetic field of induction \(\vec{B}\) exists in a direction parallel to \(P Q\). Then the force on the side \(P Q\) is (A) \(I I B\) (B) \(\frac{I l B}{2}\) (C) \(\left(\frac{\Pi B}{2}\right) \sqrt{3}\) (D) Zero

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is (A) \(\left(\frac{R_{1}}{R_{2}}\right)^{1 / 2}\) (B) \(\frac{R_{2}}{R_{1}}\) (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) (D) \(\left(\frac{R_{1}}{R_{2}}\right)\)

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then \([\mathbf{2 0 0 5}]\) (A) its velocity will increase. (B) its velocity will decrease. (C) it will turn towards left of direction of motion. (D) it will turn towards right of direction of motion.

Two long parallel wires \(P\) and \(Q\) are held perpendicular to the plane of the paper at a separation of \(5 \mathrm{~m}\). If \(P\) and \(Q\) carry currents of \(2.5 \mathrm{~A}\) and \(5 \mathrm{~A}\), respectively, in the same direction, then the magnetic field at a point midway between \(P\) and \(Q\) is (A) \(\frac{\mu_{0}}{\pi}\) (B) \(\frac{\sqrt{3} \mu_{0}}{\pi}\) (C) \(\frac{\mu_{0}}{2 \pi}\) (d) \(\frac{3 \mu_{0}}{2 \pi}\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Kinematics Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Waves Physics

Read Explanation

Oscillations

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.