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Chapter 29: Problem 60

A 25.0-cm-long metal rod lies in the \(xy\)-plane and makes an angle of 36.9\(^\circ\) with the positive \(x\)-axis and an angle of 53.1\(^\circ\) with the positive \(y\)-axis. The rod is moving in the \(+x\)-direction with a speed of 6.80 m/s. The rod is in a uniform magnetic field \(\overrightarrow{B} =\) (0.120 T)\(\hat{\imath}\) - (0.220 T)\(\hat{\jmath}\) - (0.0900 T)\(\hat{k}\). (a) What is the magnitude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

Short Answer

Expert verified
(a) 0.407 V; (b) Left end of the rod is at higher potential.

Step by step solution

01

Determine Rod's Direction Vector

The rod lies in the xy-plane, making angles of 36.9掳 with the x-axis and 53.1掳 with the y-axis. We can calculate a unit direction vector \(\hat{n}\) for the rod using these angles:\[ \hat{n} = \cos(36.9^\circ) \hat{\imath} + \cos(53.1^\circ) \hat{\jmath} \]Calculating the cosine values, we have:\[ \hat{n} = 0.8 \hat{\imath} + 0.6 \hat{\jmath} \]
02

Determine Velocity Vector of Rod

The rod is moving in the positive x-direction with a speed of 6.80 m/s. Therefore, the velocity vector \(\overrightarrow{v}\) of the rod is:\[ \overrightarrow{v} = 6.8 \hat{\imath} \]
03

Calculate Cross Product of Velocity and Magnetic Field

The emf induced in the rod can be found using the equation \( \mathcal{E} = Blv \sin(\theta) \), where \( B \) is the magnetic field, \( l \) is the length of the rod, and \( v \) is the velocity. First, we find the cross product \( \overrightarrow{v} \times \overrightarrow{B} \). Here, \(\overrightarrow{B} = 0.120 \hat{\imath} - 0.220 \hat{\jmath} - 0.0900 \hat{k}\):\[ \overrightarrow{v} \times \overrightarrow{B} = \left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \ 6.8 & 0 & 0 \ 0.120 & -0.220 & -0.0900 \end{array} \right| = (0)\hat{\imath} + (0.612)\hat{\jmath} + (1.496)\hat{k} \]
04

Calculate Induced EMF Magnitude

The magnitude of the induced emf is given by:\[ \mathcal{E} = |\overrightarrow{v} \times \overrightarrow{B}| \cdot l \cdot \hat{n} \]The magnitude of \( \overrightarrow{v} \times \overrightarrow{B} \) calculated previously is:\[ |\overrightarrow{v} \times \overrightarrow{B}| = \sqrt{0.612^2 + 1.496^2} \approx 1.626 \]The induced emf is:\[ \mathcal{E} = 1.626 \times 0.25 = 0.4065 \text{ V} \]
05

Determine Which End is at High Potential

Using the right-hand rule, with v pointing in the +x direction and \(B\) pointing into the recognizably negative z- or y-axis direction, the force on positive charges will be such that the higher potential end of the rod lies further in the direction of the rotated direction vectors. Therefore, the induced current will flow opposite \(\hat{n}\). In this case, the left end (relative to the original position moving in the x-direction) is at higher potential.

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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 vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. It is denoted by the symbol \( \overrightarrow{B} \), and in this exercise, it is given as \(0.120 \; \text{T} \; \hat{\imath} - 0.220 \; \text{T} \; \hat{\jmath} - 0.0900 \; \text{T} \; \hat{k} \). Magnetic fields can exert a force on moving charges and are responsible for the torque that acts on magnetic dipoles.
  • The unit of magnetic field strength is Tesla (T).
  • Magnetic fields can be created by electric currents, and vary with the strength and configuration of the current.
  • They are invisible but can be detected using tools like a compass, where the needle aligns with the magnetic field lines.
In this problem, the magnetic field interacts with the moving rod, leading to the generation of an electromotive force (EMF), a phenomenon described by Faraday's law of electromagnetic induction.
Cross Product
The cross product is a mathematical operation that takes two vectors and produces a third vector perpendicular to the plane containing the initial vectors. In the context of this problem, the cross product of the rod's velocity vector \( \overrightarrow{v} \) and the magnetic field \( \overrightarrow{B} \) is critical in determining the induced electromotive force (EMF) on the rod.
  • The cross product of two vectors \( \overrightarrow{A} \times \overrightarrow{B} \) is calculated using a determinant involving the unit vectors \( \hat{\imath}, \hat{\jmath}, \) and \( \hat{k} \).
  • The magnitude of the cross product is given by \( |\overrightarrow{A}| |\overrightarrow{B}| \sin(\theta) \), where \( \theta \) is the angle between the two vectors.
  • In this problem, the calculation gives us a vector, which when magnified by the length of the rod and the unit direction vector, gives the EMF.
Understanding the cross product helps visualize how the velocity of the rod and the magnetic field interact perpendicularly to create an EMF on the rod.
Unit Direction Vector
The unit direction vector is a vector that specifies the direction of an object and has a magnitude of 1. It is used to simplify calculations by eliminating scalar magnitudes when determining direction. In this exercise, the unit direction vector \( \hat{n} \) of the rod was determined from its angular position in the xy-plane.
  • It is calculated by normalizing a vector, dividing each component by the vector's overall magnitude.
  • For this problem, the angles provided allow the determination of \( \hat{n} = 0.8 \hat{\imath} + 0.6 \hat{\jmath} \).
  • This vector is crucial in determining how the induced electric field lines align with the rod's length, affecting the EMF calculation.
By understanding and computing the unit direction vector, students can accurately model how angles relate to vector components in spatial problems.
Kinematics in Physics
Kinematics in physics refers to the study of motion without considering the forces that cause it. It involves parameters such as velocity, displacement, and acceleration. In this exercise, kinematics is represented by the velocity of the rod moving in the positive x-direction with a speed of 6.80 m/s.
  • Velocity is a vector quantity, meaning it has both magnitude (speed) and direction.
  • Displacement vectors can be resolved into orthogonal components, simplifying analysis in physics problems.
  • Kinematics allows us to determine the potential influence of various motions within physical fields such as a magnetic field.
In this scenario, kinematics allows us to model how the rod's movement through the magnetic field changes its electromagnetic properties, leading to the generation of an induced EMF, further illustrating the interplay between motion and electromagnetic phenomena.

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Most popular questions from this chapter

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 cm\(^2\) is rotated in 0.040 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 6.0 \(\times\) 10\(^{-5}\) T. (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?

A closely wound search coil (see Exercise 29.3) has an area of 3.20 cm\(^2\), 120 turns, and a resistance of 60.0 \(\Omega\). It is connected to a charge- measuring instrument whose resistance is 45.0 \(\Omega\). When the coil is rotated quickly from a position parallel to a uniform magnetic field to a position perpendicular to the field, the instrument indicates a charge of 3.56 \(\times\) 10\(^{-5}\) C. What is the magnitude of the field?

A square, conducting, wire loop of side L, total mass m, and total resistance R initially lies in the horizontal xy-plane, with corners at (\(x, y, z\)) = (0, 0, 0), (0, \(L\), 0), (\(L\), 0, 0), and (\(L, L\), 0). There is a uniform, upward magnetic field \(\overrightarrow{B}\) = B\(\hat{k}\) in the space within and around the loop. The side of the loop that extends from (0, 0, 0) to (\(L\), 0, 0) is held in place on the \(x\)-axis; the rest of the loop is free to pivot around this axis. When the loop is released, it begins to rotate due to the gravitational torque. (a) Find the \(net\) torque (magnitude and direction) that acts on the loop when it has rotated through an angle \(\phi\) from its original orientation and is rotating downward at an angular speed \(\omega\). (b) Find the angular acceleration of the loop at the instant described in part (a). (c) Compared to the case with zero magnetic field, does it take the loop a longer or shorter time to rotate through 90\(^\circ\) ? Explain. (d) Is mechanical energy conserved as the loop rotates downward? Explain.

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal \(xy\)-plane is located in a region of uniform magnetic field. A field of 1.5 T is directed along the positive z-direction, which is upward. (a) If the loop is removed from the field region in a time interval of 2.0 ms, find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

A 0.650-m-long metal bar is pulled to the right at a steady 5.0 m/s perpendicular to a uniform, 0.750 T magnetic field. The bar rides on parallel metal rails connected through a 25.0-\(\Omega\) resistor (\(\textbf{Fig. E29.30}\)), so the apparatus makes a complete circuit. Ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit by using (i) the magnetic force on the charges in the moving bar; (ii) Faraday's law; (iii) Lenz's law. (c) Calculate the current through the resistor.
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