/*! 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 119 Figure \(16.66\) shows four rods... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 119

Figure \(16.66\) shows four rods having \(\lambda=0.5 \Omega / \mathrm{m}\) resistance per unit length. The arrangement is kept in a magnetic field of constant magnitude \(B=2 \mathrm{~T}\) and directed perpendicular to the plane of the figure and directed inwards. Initially, the rods form a square of side length \(\ell=15 \mathrm{~m}\) as shown. Now each wire starts moving with constant velocity \(v=5 \mathrm{~m} / \mathrm{s}\) towards opposite wire. Find the force required in newton on each wire to keep its velocity constant at \(t=1 \mathrm{~s}\).

Short Answer

Expert verified
The force required to keep the velocity constant for each rod at t = 1s is 0 N.

Step by step solution

01

Calculation of the new side length of square at t=1s

Since rods are moving towards each other with a constant velocity v = 5 m/s and we need to find the force required to keep their velocity constant at t = 1s, first let's find the new side length of the square after 1 second. As the rods are moving towards each other, in 1 second, the square side will decrease by 5 meters. New side length = Initial side length - 2 * (Distance travelled in 1 sec) New side length = 15 m - (2*(5 m)) = 5 m
02

Calculation of the magnetic flux and induced EMF

We will now calculate the magnetic flux, and then use Faraday's law to find the induced EMF. The magnetic flux (桅) is given by the product of magnetic field (B), area (A) and the cosine of the angle between the magnetic field and area vector. In this case, the angle is 90掳. 桅 = BA cos(90掳) 桅 = 2 T * (5 m * 5 m) * cos(90掳) 桅 = 50 Tm虏 * 0 = 0 Since the magnetic flux is not changing, the induced EMF is zero.
03

Calculation of the current flow

Since the induced EMF is zero, no current flows through the rod due to the magnetic field. Therefore, there is no magnetic force acting on the rods.
04

Calculation of the force required to keep velocity constant

Since there is no magnetic force acting on the rods, the only force that needs to be applied to maintain the constant velocity is the resistive force due to the resistance of the rods. As there is no current flow through the rods, the resistive force is also zero. Thus, the force required to keep the velocity constant for each rod at t = 1s is 0 N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Flux
Magnetic flux is a fundamental concept in electromagnetism, describing the number of magnetic field lines passing through a given area. It is represented by the Greek letter 桅 (phi). To determine its value, we use the formula:
  • 桅 = B 脳 A 脳 cos(胃)
where: - B is the magnetic field strength, - A is the area through which the field lines pass, - 胃 is the angle between the magnetic field and the normal to the surface.
In the provided exercise, the magnetic flux calculation showed that when the side length of the square reduces to 5 m, the perpendicular magnetic flux through the plane is zero. This is because the cosine of 90 degrees leads to zero flux, indicating no magnetic field lines pass through the reduced area.
Faraday's Law
Faraday's Law of electromagnetic induction is a cornerstone of electromagnetism. It explains how a voltage or electromotive force (EMF) is induced in a conductor when it is exposed to a changing magnetic field.
The formula for Faraday's Law is:
  • 蔚 = -d桅/dt
where: - 蔚 is the induced EMF, - d桅/dt is the rate of change of magnetic flux
In our exercise, despite the movement of the rods, no EMF is induced. This outcome is because the magnetic flux remains constant (zero change in motion), therefore not fulfilling the requirement for a changing magnetic flux as stated in Faraday's Law.
Induced EMF
Induced EMF refers to the voltage generated in a closed conducting loop when it experiences a change in magnetic flux. It is central to motors and generators' working phenomena.
Induced EMF can occur due to:
  • Change in the magnetic field strength,
  • Change in the area of the conducting loop within the magnetic field,
  • Change in orientation of the loop relative to the magnetic field.
In the exercise scenario, although the wires move, the magnetic flux through the circuit is constant, resulting in zero induced EMF. This conclusion shows the importance of flux change for EMF production and why no current was induced.
Resistance per Unit Length
Resistance per unit length, denoted as 位, is essential for understanding how a material鈥檚 resistance scales with length. It is calculated using:
  • Resistance, R = 位 脳 L
where: - 位 is the resistance per unit length, - L is the length of the conducting material.
In the original problem, each rod has a resistance of 0.5 惟/m. Despite the immense importance of resistance determining current flow, the exercise confirmed zero current and, consequently, zero resistive forces since neither a magnetic force nor an induced EMF was present amongst the rods at t = 1s.

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Most popular questions from this chapter

In a uniform magnetic field of induction \(B\), a wire in the form of a semicircle of radius \(r\) rotates about the diameter of the circle with an angular frequency \(\omega\). The axis of rotation is perpendicular to the field. If the total resistance of the circuit is \(R\), the mean power generated per period of rotation is (A) \(\frac{(B \pi r \omega)^{2}}{2 R}\) (B) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{8 R}\) (C) \(\frac{B \pi r^{2} \omega}{2 R}\) (D) \(\frac{\left(B \pi r \omega^{2}\right)^{2}}{8 R}\)

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

In a uniform magnetic field of induction \(B\), a wire in the form of semicircle of radius \(r\) rotates about the diameter of the circle with angular frequency \(\omega\). If the total resistance of the circuit is \(R\), the mean power generated per period of rotation is (A) \(\frac{B \pi r^{2} \omega}{2 R}\) (B) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{2 R}\) (C) \(\frac{(B \pi r \omega)^{2}}{2 R}\) (D) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{8 R}\)

Induced electric field (in volt/meter) at the circumference of ring at the instant ring starts toppling is (A) \(\frac{10}{\pi}\) (B) \(\frac{20}{\pi}\) (C) \(\frac{5}{\pi}\) (D) \(\frac{25}{\pi}\)

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) \(g e^{-4}\)
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