Problem 60 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: Determine Rod's Direction Vector

The rod lies in the xy-plane, making angles of 36.9° with the x-axis and 53.1° 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:\[ \hat{n} = 0.8 \hat{\imath} + 0.6 \hat{\jmath} \]

Step 2: 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:\[ \overrightarrow{v} = 6.8 \hat{\imath} \]

Step 3: 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, \(\overrightarrow{B} = 0.120 \hat{\imath} - 0.220 \hat{\jmath} - 0.0900 \hat{k}\):\[ \overrightarrow{v} \times \overrightarrow{B} = \left| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \ 6.8 & 0 & 0 \ 0.120 & -0.220 & -0.0900 \end{array} \right| = (0)\hat{\imath} + (0.612)\hat{\jmath} + (1.496)\hat{k} \]

Step 4: 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:\[ \mathcal{E} = 1.626 \times 0.25 = 0.4065 \text{ V} \]

Step 5: 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.

Key Concepts

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.