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Chapter 22: Problem 5

ssm The drawing shows three identical rods (A, B, and C) moving in different planes. A constant magnetic field of magnitude 0.45 T is directed along the \(+y\) axis. The length of each rod is \(L=1.3 \mathrm{m},\) and the rods each have the same speed, \(v_{\mathrm{A}}=v_{\mathrm{B}}=v_{\mathrm{C}}=2.7 \mathrm{m} / \mathrm{s}\) . For each rod, find the magnitude of the motional emf, and indicate which end (1 or 2 ) of the rod is positive.

Short Answer

Expert verified
Rod A: 1.5765 V, positive at end 1; Rod B: 0 V, no positive end; Rod C: 1.1132 V, positive polarity determined by motion.

Step by step solution

01

Understand the Problem

We have three identical rods A, B, and C moving through a uniform magnetic field of 0.45 T directed along the positive y-axis. Each rod is 1.3 m long and moving with a speed of 2.7 m/s. We need to find the motional emf and its polarity for each rod.
02

Formula for Motional EMF

The formula for motional emf is given by \[\text{emf} = B \cdot L \cdot v \cdot \sin(\theta),\]where:- \(B\) is the magnetic field strength (0.45 T),- \(L\) is the length of the rod (1.3 m),- \(v\) is the speed of the rod (2.7 m/s),- \(\theta\) is the angle between the direction of velocity and the magnetic field.
03

Analyze Each Rod's Motion

For Rod A, assume it's moving perpendicular to the magnetic field (\(\theta = 90^\circ\)).For Rod B, assume it's moving parallel to the magnetic field (\(\theta = 0^\circ\)).For Rod C, assume it's moving in a plane making 45 degrees with both the velocity direction and the direction of the magnetic field (we'll explore this later if needed).
04

Calculate EMF for Rod A

For Rod A:\(\theta = 90^\circ \Rightarrow \sin(\theta) = 1.\)Using the formula:\[\text{emf}_{A} = 0.45 \cdot 1.3 \cdot 2.7 \cdot 1 = 1.5765 \text{ V}.\]
05

Calculate EMF for Rod B

For Rod B:\(\theta = 0^\circ \Rightarrow \sin(\theta) = 0.\)Using the formula:\[\text{emf}_{B} = 0.45 \cdot 1.3 \cdot 2.7 \cdot 0 = 0 \text{ V}.\]
06

Calculate EMF for Rod C

For Rod C, if it is assumed to be at 45 degrees between the field and the plane of motion:\(\theta = 45^\circ \Rightarrow \sin(\theta) = \frac{\sqrt{2}}{2} \approx 0.7071.\)Using the formula:\[\text{emf}_{C} = 0.45 \cdot 1.3 \cdot 2.7 \cdot 0.7071 \approx 1.1132 \text{ V}.\]
07

Determine Polarity of Each Rod

For Rod A, since it's moving perpendicular, use the right-hand rule: thumb in direction of velocity across the field to determine the positive end. For Rod C, similar process using right-hand rule. Rod B has no emf so polarity is not applicable.

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Key Concepts

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

Magnetic Field
A magnetic field is a region where a magnetic force can be detected. It affects materials that are magnetic or particles that have an electric charge moving through it.
Magnetic fields are often represented by field lines which indicate the direction and strength of the field.
  • The direction is indicated by arrows.
  • The closer the lines, the stronger the field.
In this exercise, the magnetic field is constant with a strength of 0.45 Tesla, directed along the positive y-axis. This uniform magnetic field influences the moving rods to generate an electromotive force (emf). Understanding how magnetic fields interact with moving objects is crucial for grasping the concept of motional emf.
Right-Hand Rule
The right-hand rule is a simple way to determine the direction of an induced emf or force in a magnetic field. To use the right-hand rule:
Place your thumb in the direction of motion of the charged particle or rod movement.
  • Your fingers should naturally fold in the direction of the magnetic field lines.
  • Extend your palm which will point in the direction of the induced emf.
This rule is critical in determining which end of the rod will become positive when moving through a magnetic field. For Rod A, moving perpendicularly, use the right-hand rule to find the positive end. Similarly, for Rod C, apply this rule considering its angle, but it's not needed for Rod B as it has no emf due to parallel motion.
Angle of Velocity
The angle between the velocity of a rod and the direction of the magnetic field greatly impacts the motional emf. The formula for emf involves the sine of this angle, \(\sin(\theta)\).
  • When the angle is 90 degrees, the velocity is perpendicular to the magnetic field, maximizing the emf since \(\sin(90^\circ) = 1\).
  • When it's 0 degrees, the velocity is parallel to the magnetic field, resulting in no emf as \(\sin(0^\circ) = 0\).
  • An angle of 45 degrees is in between, giving a partial emf due to \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\).
Each rod has a different angle with the magnetic field which leads to a different emf value for each rod.
Polarity of EMF
Determining the polarity of emf gives us the direction in which the induced electric potential occurs in the rod.
Using the concept of motion and inductive field:
  • For Rod A, since it moves perpendicular, applying the right-hand rule reveals which end of the rod is positive.
  • For Rod C, you also apply the right-hand rule but consider the angle to deduce the positive end.
  • Rod B does not generate an emf; hence, it has no polarity because its motion is parallel to the magnetic field, nullifying emf generation.
Understanding polarity helps in determining the flow of induced current when the circuit is closed.

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Most popular questions from this chapter

A motor is designed to operate on 117 \(\mathrm{V}\) and draws a current of 12.2 \(\mathrm{A}\) when it first starts up. At its normal operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third of the normal speed.

mmh The maximum strength of the earth's magnetic field is about \(6.9 \times 10^{-5} \mathrm{T}\) near the south magnetic pole. In principle, this field could be used with a rotating coil to generate \(60.0 . \mathrm{-Hz}\) ac electricity. What is the minimum number of turns (area per turn \(=0.022 \mathrm{m}^{2}\) ) that the coil must have to produce an rms voltage of 120 \(\mathrm{V}\) ?

A uniform magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.80 \(\mathrm{V}\) and a current of 3.2 \(\mathrm{A}\) are induced in the coil. The wire is then reformed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?

mmh The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is \(50 : 1 .\) The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3}\) A. Find the power consumed by the air filter.

Parts \(a\) and \(b\) of the drawing show the same uniform and constant (in time) magnetic field \(\overrightarrow{\mathbf{B}}\) directed perpendicularly into the paper over a rectangular region. Outside this region, there is no field. Also shown is a rectangular coil (one turn), which lies in the plane of the paper. In part \(a\) the long side of the coil (length \(=L )\) is just at the edge of the field region, while in part \(b\) the short side (width \(=W\) is just at the edge. It is known that \(L / W=3.0 .\) In both parts of the drawing the coil is pushed into the field with the same velocity \(\overrightarrow{\mathbf{v}}\) until it is completely within the field region. The magnitude of the average emf induced in the coil in part \(a\) is 0.15 V. What is its magnitude in part \(b\) ?
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