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A 25.0 -cm-long metal rod lies in the \(x y\) -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 \(\mathrm{m} / \mathrm{s}\) . The rod is in a uniform magnetic field \(\vec{B}=(0.120 \mathrm{T}) \hat{i}-(0.220 \mathrm{T}) \hat{J}-(0.0900 \mathrm{T}) \hat{k}\) (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) 0.3234 V (b) Right-hand end at higher potential in +x-direction.

Step by step solution

01

Calculate the length component along the x-axis

The rod makes an angle of \(36.9^\circ\) with the \(x\)-axis. We calculate the length component of the rod along the \(x\)-axis using the cosine of this angle: \[ L_x = L \cos(36.9^\circ) = 25.0\ cm \times \cos(36.9^\circ) \approx 20.0\ cm \] where \(L = 25.0\ cm\). Convert this to meters: \(L_x = 0.20\ m\).
02

Calculate the velocity vector

Since the rod is moving in the \(+x\)-direction, the velocity vector \(\vec{v}\) can be expressed as:\[ \vec{v} = (6.80 \mathrm{m/s})\hat{i} \] This means the velocity only has a non-zero component in the \(x\)-direction.
03

Use the formula for EMF in a moving conductor

The emf \(\varepsilon\) induced in the rod is given by \[ \varepsilon = \left| \vec{v} \times \vec{B} \right| L \] Calculate the cross product \(\vec{v} \times \vec{B}\):1. \[ \vec{v} = (6.80 \mathrm{m/s})\hat{i} \]2. \[ \vec{B} = (0.120 \mathrm{T}) \hat{i} - (0.220 \mathrm{T}) \hat{j} - (0.0900 \mathrm{T}) \hat{k} \]The cross product gives:\[ \vec{v} \times \vec{B} = |\hat{i}\ \hat{j}\ \hat{k}|\begin{vmatrix}6.80 & 0 & 0 \0.120 & -0.220 & -0.090\end{vmatrix} \]Solving the determinant, we get \[ \vec{v} \times \vec{B} = (0)(-0.090) - (0)(-0.220)\hat{i} - [(6.80)(-0.090) - (0)(0.120)]\hat{j} + [(6.80)(-0.220) - (0)(0.120)]\hat{k} \] which simplifies to: \[ \vec{v} \times \vec{B} = 0\hat{i} + 0.612\hat{j} - 1.496\hat{k} \]
04

Calculate the magnitude of the cross product

Calculate the magnitude of \(\vec{v} \times \vec{B}\):\[ |\vec{v} \times \vec{B}| = \sqrt{(0)^2 + (0.612)^2 + (-1.496)^2}\]\[ = \sqrt{0 + 0.374544 + 2.238016} = \sqrt{2.61256} \approx 1.617 \mathrm{T}\cdot \mathrm{m/s} \].
05

Calculate the induced EMF

Now, find the induced emf using:\[ \varepsilon = |\vec{v} \times \vec{B}| \times L_x \]\[ \varepsilon = 1.617 \times 0.20 = 0.3234 \mathrm{V} \].
06

Determine which end is at a higher potential

According to the right-hand rule, if the fingers point in the direction of velocity and curl toward the magnetic field, the thumb points in the direction of positive emf. Here, the induced electric field will point towards the \(-y\)-axis, making the end of the rod that originally lied closer to -y-axis of 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 an invisible force field created by electric currents and magnets. It is depicted by field lines that run from the north to the south pole. These fields are crucial for understanding how charged particles move in space. In physics, the magnetic field vector is often denoted as \( \vec{B} \). This vector has three components corresponding to its orientation in three-dimensional space:
  • \( B_x \): the component along the x-axis
  • \( B_y \): the component along the y-axis
  • \( B_z \): the component along the z-axis
For our exercise, the magnetic field vector is given by \( \vec{B} = (0.120 \mathrm{T}) \, \hat{i} - (0.220 \mathrm{T}) \, \hat{j} - (0.0900 \mathrm{T}) \, \hat{k} \). Here, each component indicates the strength and direction of the field in that respective axis. Understanding the magnetic field's orientation helps us analyze how the field interacts with moving charges, such as those in a metal rod.
Cross Product Calculation
In vector mathematics, the cross product is a method to find a vector that is orthogonal to two given vectors. This is particularly important in physics for calculating induced emf, as it gives us a new vector that helps describe the interaction of velocity and magnetic field vectors. Given two vectors \( \vec{a} \) and \( \vec{b} \), the cross product \( \vec{a} \times \vec{b} \) is computed using the determinant of a matrix involving unit vectors:
  • The top row contains the unit vectors \( \hat{i}, \hat{j}, \hat{k} \).
  • The second row contains the components of \( \vec{a} \).
  • The third row contains the components of \( \vec{b} \).
For the exercise, calculating \( \vec{v} \times \vec{B} \) is accomplished by:\[\vec{v} \times \vec{B} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 6.80 & 0 & 0 \ 0.120 & -0.220 & -0.090 \end{array}| \]This determinant expands to yield a new vector, showing how velocity interacts with the magnetic field across different dimensions. This operation results in a vector that is crucial for determining the emf.
Right-Hand Rule
The right-hand rule is a mnemonic tool used in vector mathematics, specifically for determining directions in cross product calculations. To apply the right-hand rule, point your thumb in the direction of the first vector (velocity), and curl your fingers toward the second vector (magnetic field). The direction your palm faces or your thumb points, after executing this maneuver, shows the direction of the resultant vector.
  • Thumb: direction of velocity \( \vec{v} \)
  • Fingers: curl in the direction of the magnetic field \( \vec{B} \)
  • Palm/thumb: direction of induced emf or force
In this exercise, this rule helps determine the orientation of the induced electromotive force (emf) in the rod. By using the right-hand rule, we can visualize and confirm that the induced electric field points towards the \(-y\)-axis, indicating which end of the rod is at a higher potential.

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Most popular questions from this chapter

CALC An airplane propeller of total length \(L\) rotates around its center with angular speed \(\omega\) in a magnetic field that is perpendicular to the plane of rotation. Modeling the propeller as a thin, uniform bar, find the potential difference between (a) the center and either end of the propeller and (b) the two ends. (c) If the field is the earth's field of 0.50 \(\mathrm{G}\) and the propeller turns at 220 \(\mathrm{rpm}\) and is 2.0 \(\mathrm{m}\) long, what is the potential difference between the middle and either end? It this large enough to be concerned about?

Back emf. A motor with a brush-and-commutator arrangement, as described in Example \(29.4,\) has a circular coil with radius 2.5 \(\mathrm{cm}\) and 150 turns of wire. The magnetic field has magnitude \(0.060 \mathrm{T},\) and the coil rotates at 440 \(\mathrm{rev} / \mathrm{min.}\) (a) What is the maximum emf induced in the coil? (b) What is the average back emf?

Motional emfs in Transportation. Airplanes and trains move through the earth's magnetic field at rather high speeds, so it is reasonable to wonder whether this field can have a substantial effect on them. We shall use a typical value of 0.50 G for the earth's field (a) The French TGV train and the Japanese "bullet train" reach speeds of up to 180 mph moving on tracks about 1.5 \(\mathrm{m}\) apart. At top speed moving perpendicular to the earth's magnetic field, what potential difference is induced across the tracks as the wheels roll? Does this seem large enough to produce noticeable effects? (b) The Boeing \(747-400\) aircraft has a wingspan of 64.4 \(\mathrm{m}\) and a cruising speed of 565 mph. If there is no wind blowing (so that this is also their speed relative to the ground), what is the maximum potential difference that could be induced between the opposite tips of the wings? Does this seem large enough to cause problems with the plane?

A closely wound rectangular coil of 80 turns has dimensions of 25.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) . The coll is rotated from a position where it makes an angle of \(37.0^{\circ}\) with a magnetic field of 1.10 \(\mathrm{T}\) to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf induced in the coil?

Terminal Speed. A conducting rod with length \(L\) mass \(m,\) and resistance \(R\) moves without friction on metal rails as shown in Fig. \(29.11 .\) A uniform magnetic field \(\vec{B}\) is directed into the plane of the figure. The rod starts from rest and is acted on by a constant force \(\vec{\boldsymbol{F}}\) directed to the right. The rails are infinitely long and have negligible resistance. (a) Graph the speed of the rod as a function of time. (b) Find an expression for the terminal speed (the speed when the acceleration of the rod is zero).

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