/*! 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 52 Current \(i\) is carried in a wi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 52

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 maximum magnitude of torque in a given magnetic field \(B\) will be \(\tau_\text{max} = \frac{L^{2} i B}{4 \pi}\).

Step by step solution

01

The magnetic moment \(\boldsymbol{碌}\) can be calculated using the formula: \[\boldsymbol{碌} = NIA\hat{n}\] where \(N\) is the number of turns in the coil, \(I\) is the current, \(A\) is the area of the coil, and \(\hat{n}\) is the normal unit vector. In this case, we have only one turn, so \(N = 1\). To find the area, we need to find the radius of the circular coil formed by the wire. The length of the wire, \(L\), is equal to the circumference of the circular coil, so \[L = 2 \pi r\] Solving for the radius, we get: \[r = \frac{L}{2\pi}\] Now we can find the area of the circular coil: \[A = \pi r^2 = \pi \left( \frac{L}{2\pi} \right)^2\] The magnetic moment becomes: \[\boldsymbol{碌} = I \pi \left(\frac{L}{2\pi}\right)^2 \hat{n} = \frac{I L^2}{4\pi} \hat{n}\] #Step 2: Find the torque on the circular coil#

Use the formula for torque \(\boldsymbol{\tau} = \boldsymbol{碌} \times \boldsymbol{B}\) and the magnetic moment we found in step 1: \[\boldsymbol\tau = \left(\frac{IL^2}{4\pi}\hat{n}\right) \times (B\hat{k})\] Now, we want to find the maximum torque. The maximum torque will be achieved when the angle between \(\boldsymbol\mu\) and \(\boldsymbol B\) is 90掳, as the cross product formula for torque involves a sine term: \[\boldsymbol\tau_\text{max} = \frac{IL^2}{4\pi} B\] Comparing our expression for maximum torque with the given options, we find that it matches option (C): \[\boxed{\tau_\text{max} = \frac{L^{2} i B}{4 \pi}}\]

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

Two straight long conductors \(A O B\) and \(C O D\) are perpendicular to each other and carry currents \(i_{1}\) and \(i_{2}\). 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 \(A C B D\) 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) \(\frac{\mu_{0}}{2 \pi a} \frac{i_{1} i_{2}}{\left(i_{1}+i_{2}\right)}\)

Two identical wires \(A\) and \(B\), each of length \(\cdot \ell^{\prime}\), carry the same current I. Wire \(A\) is bent into a circle of radius \(R\) and wire \(B\) is bent to form a square of side \({ }^{\prime} \mathrm{a}\) '. If \(B_{\mathrm{A}}\) and \(B_{\mathrm{B}}\) are the values of magnetic field at the centres of the circle and square respectively, then the ratio \(\frac{B_{A}}{B_{B}}\) is (A) \(\frac{\pi^{2}}{16 \sqrt{2}}\) (B) \(\frac{\pi^{2}}{16}\) (C) \(\frac{\pi^{2}}{8 \sqrt{2}}\) (D) \(\frac{\pi^{2}}{8}\)

Two circular coils of radii \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) carry equal currents of \(2 \mathrm{~A}\). The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes and their centres coincide. Magnitude of magnetic field at the common centre of coils is, (A) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in same direction. (B) \(4 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction. (C) zero, if currents in the coils are in opposite direction. (D) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction.

A proton with a speed \(u\) along the positive \(x\)-axis at \(y=0\) enters a region of uniform magnetic field \(B=-B_{0} \hat{k}\) which exists to the right of \(y\)-axis. The proton exits from the region after some time with the speed \(v\) at ordinate \(y\), then (A) \(v>u, y<0\) (B) \(v=u, y>0\) (C) \(v>u, y>0\) (D) \(v=u, y<0\)

Which of the following is true? (A) Diamagnetism is temperature dependent (B) Paramagnetism is temperature dependent (C) Paramagnetism is temperature independent (D) None of the above
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Wave Optics

Read Explanation

Astrophysics

Read Explanation

Electromagnetism

Read Explanation

Electrostatics

Read Explanation

Radiation

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.