/*! 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 30 A conducting rod \(P Q\) is movi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 30

A conducting rod \(P Q\) is moving parallel to \(x-z\)-plane in a uniform magnetic field directed in the positive \(y\)-direction. The end \(P\) of the rod will become (A) Sometime positive and sometime negative (B) Positive (C) Neutral (D) Negative

Short Answer

Expert verified
The end \(P\) of the rod will become positive.

Step by step solution

01

Understand the theoretical concept

Understand the principle of electromagnetic induction. According to Faraday's law of electromagnetic induction, a change in magnetic flux through a loop of wire induces an electromotive force (emf) in the wire. The induced emf creates an electric field causing the charge carriers (electrons) to move. Consider the direction of motion of these charge carriers due to the induced emf.
02

Applying Right Hand Rule on Conductor

Apply the right-hand rule to the given system. According to this rule, if the thumb points in the direction of the motion of the conductor and the fingers towards the direction of the magnetic field, then the direction of the induced current (and hence the direction of the induced emf) will be in the direction in which the palm faces. In this case, as the conductor moves in the \(x-z\) plane and the magnetic field is upward (positive y-direction), the induced emf will be from Q to P.
03

Determining the Charge at end P

The direction of the induced emf determines the direction of the movement of charges. Thus, based on step 2, the induced emf direction is from Q to P, causing the electrons (which are negatively charged) to move towards end Q. This leaves end P with a deficiency of negatively charged electrons, making it positively charged.

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 coil having \(n\) turns and resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 R \Omega\). This combination is moved in time \(t\) seconds from a magnetic field \(W_{1}\) weber to \(W_{2}\) weber. The induced current in the circuit is (A) \(-\frac{\left(W_{2}-W_{1}\right)}{R n t}\) (B) \(-\frac{n\left(W_{2}-W_{1}\right)}{5 R t}\) (C) \(-\frac{\left(W_{2}-W_{1}\right)}{5 R n t}\) (D) \(-\frac{n\left(W_{2}-W_{1}\right)}{R t}\)

A conducting rod of length \(2 \ell\) is rotating with constant angular speed \(\omega\) about its perpendicular bisector. A uniform magnetic field \(\vec{B}\) exists parallel to the axis of rotation. The EMF induced between two ends of the rod is (A) \(B \omega \ell^{2}\) (B) \(\frac{1}{2} B \omega \ell^{2}\) (C) \(\frac{1}{8} B \omega \ell^{2}\) (D) Zero

A magnetic field \(\vec{B}=\left(\frac{B_{0} y}{a}\right) \hat{k}\) is into the paper in the \(+z\) direction. \(B_{0}\) and \(a\) are positive constants. A square loop \(E F G H\) of side \(a\), mass \(m\), and resistance \(R\), in \(x-y\) plane, starts falling under the influence of gravity. Assume \(x\)-axis is horizontal and \(y\) is vertically downward. Acceleration of the loop when its speed is half of its terminal speed is (A) \(\frac{g}{2}\) (B) \(g e^{-2}\) (C) \(g e^{-1 / 2}\) (D) \(g e^{-4}\)

If a coil of metal wire is kept stationary in a non-uniform magnetic field, (A) An EMF and current are both induced in the coil (B) A current but no EMF is induced in the coil (C) An EMF but no current is induced in the coil (D) Neither EMF nor current is induced in the coil

Net force on a current carrying loop kept in uniform magnetic field is zero and the torque on the loop \(\vec{\tau}=\vec{M} \times \vec{B}\), where \(M\) and \(B\) are magnetic dipole moment and magnetic field intensity, respectively. If it is free to rotate, then it will rotates about an axis passing through its centre of mass and parallel to \(\vec{\tau}\). Potential energy of the loop is given by \(U=-\vec{M} \cdot \vec{B}\). Assume a current carrying ring with its centre at the origin and having moment of inertia \(2 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{2}\) about an axis passing through one of its diameter and magnetic moment \(\vec{M}=(3 \hat{i}-4 \hat{j}) \mathrm{Am}^{2}\). At time \(t=0\), a magnetic field \(\vec{B}=(4 \hat{i}-3 \hat{j}) T\) is switched on. Then Angular acceleration of the ring at time \(t=0\) (in \(\mathrm{rad} / \mathrm{s}^{2}\) ) is (A) 5000 (B) 1250 (C) 2500 (D) Zero
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Famous Physicists

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Scientific Method Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Rotational Dynamics

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.