/*! 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 35 A conductor of length \(l\) and ... [FREE SOLUTION] | 91Ó°ÊÓ

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A conductor of length \(l\) and mass \(m\) can slide along a pair of vertical metal guides connected by a resistor \(R .\) A uniform magnetic field of strength \(B\) normal to the plane of page is directed outwards. The steady speed of fall of rod is: (a) \(\frac{m g R}{B^{2} l^{2}}\) (b) \(\frac{m g}{B^{2} l^{2} R}\) (c) \(\frac{B^{2} l^{2}}{m g R}\) (d) \(\frac{m g B}{l^{2} R}\)

Short Answer

Expert verified
The steady speed of fall is \( \frac{mgR}{B^2l^2} \) (choice (a)).

Step by step solution

01

Understanding the Problem

We have a conductor of length \( l \) and mass \( m \) that slides along vertical metal guides connected to a resistor \( R \), with a magnetic field \( B \) perpendicular to the plane of the page. We need to find the steady speed of the rod as it falls.
02

Applying Faraday's Law of Induction

As the conductor falls with speed \( v \), it cuts through magnetic field lines, inducing an electromotive force (EMF) given by \( \text{EMF} = B \cdot l \cdot v \).
03

Calculating Induced Current

The induced EMF causes a current \( I \) to flow through the resistor, given by Ohm's Law: \( I = \frac{\text{EMF}}{R} = \frac{B \cdot l \cdot v}{R} \).
04

Analyzing the Forces on the Conductor

The downward gravitational force is \( F_{\text{gravity}} = mg \). The magnetic field exerts an upward Lorentz force given by \( F_{\text{magnetic}} = I \cdot B \cdot l = \frac{B^2 \cdot l^2 \cdot v}{R} \). For steady speed, these forces balance: \( mg = \frac{B^2 \cdot l^2 \cdot v}{R} \).
05

Solving for the Steady Speed

Rearrange the force balance equation to solve for \( v \): \( v = \frac{mgR}{B^2 \cdot l^2} \).

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Key Concepts

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

Faraday's Law
Faraday's Law is a fundamental principle in electromagnetism, describing how an electromotive force (EMF) is generated in a conductor when it experiences a changing magnetic field.
When a conductor, such as the one in our problem, moves through a magnetic field, it "cuts" through the magnetic lines of force. This change in magnetic flux induces an EMF in the conductor, proportional to the rate at which the conductor traverses the field.
Mathematically, Faraday’s Law is expressed as:
  • \( \text{EMF} = - \frac{d\Phi}{dt} \)
where \( \Phi \) is the magnetic flux through the circuit. In our specific scenario, the expression becomes \( \text{EMF} = B \cdot l \cdot v \), because we are considering a uniform magnetic field \( B \), the length of the conductor \( l \), and its velocity \( v \). Understanding this concept is crucial as it sets the stage for the induction of current in the circuit.
Lorentz Force
The Lorentz force is the force experienced by a charged particle moving through a magnetic and/or electric field. For our exercise, this translates to the force exerted on the conductor as it falls through the magnetic field.
This force plays a vital role because it opposes the gravitational force acting on the conductor, and thus affects its motion.
The Lorentz force is essentially the magnetic force in our scenario, calculated as:
  • \( F_{\text{magnetic}} = I \cdot B \cdot l \)
where \( I \) is the induced current, \( B \) is the magnetic field strength, and \( l \) is the length of the conductor.
By substituting for the current using Ohm's Law, we get \( I = \frac{B \cdot l \cdot v}{R} \).
Thus, the magnetic force becomes \( F_{\text{magnetic}} = \frac{B^2 \cdot l^2 \cdot v}{R} \). Understanding the balancing of this force with gravity is key to solving the problem.
Ohm's Law
Ohm's Law is a simple and essential concept in electronics, relating the current flowing through a conductor to the applied voltage and resistance. It is critical in determining the resultant current from the induced EMF in our exercise.
Ohm's Law can be expressed in the basic formula:
  • \( V = I \cdot R \)
where \( V \) is the voltage (or EMF in our context), \( I \) is the current, and \( R \) is the resistance.
By rearranging Ohm's Law, the induced current \( I \) in the conductor can be determined as:
  • \( I = \frac{\text{EMF}}{R} \)
In our scenario, \( \text{EMF} = B \cdot l \cdot v \), yielding \( I = \frac{B \cdot l \cdot v}{R} \). This current is crucial in analyzing the Lorentz force that acts on the conductor. Thus, understanding Ohm's Law helps in tying all elements of the problem together to find the rod's steady speed.

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Most popular questions from this chapter

The sum and the difference of self inductances of two coils are \(13 \mathrm{H}\) and \(5 \mathrm{H}\) respectively. The maximum mutual inductances of two coil is (a) \(6 \mathrm{H}\) (b) \(5 \mathrm{H}\) (c) \(\sqrt{65} \mathrm{H}\) (d) \(18 \mathrm{H}\)

The self inductance of the air cored solenoid of length \(80 \mathrm{~cm}\) and has 500 turns and its circular cross-section has diameter of \(2 \mathrm{~cm}\) is : (a) \(150.6 \mu \mathrm{H}\) (b) \(162.2 \mu \mathrm{H}\) (c) \(123.3 \mu \mathrm{H}\) (d) \(102.5 \mu \mathrm{H}\)

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A system \(S\) consists of two coils \(A\) and \(B\). The coil \(A\) have a steady current \(I\) while the coils \(B\) is su \(_{1}\) - nded near by as shown in figure. Now the system is hc:,\(\quad\) (a) the two coils show attraction (b) the two coils show repulsion (c) there is no change in the position of the two coils (d) induced currents are not possible in coil \(B\)

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