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A \(25.0-\mathrm{cm}\) -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 \(\overrightarrow{\boldsymbol{B}}=(0.120 \mathrm{~T}) \hat{\imath}-(0.220 \mathrm{~T}) \hat{\jmath}-(0.0900 \mathrm{~T}) \hat{\boldsymbol{k}} .\) (a) What is the magni- tude 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
The magnitude of the emf induced in the rod is 0.120 mV. In the sketch, the lower end of the rod (near the origin of axes) is at a higher potential.

Step by step solution

01

Find the velocity vector

The velocity of the rod is given as \(6.80 \mathrm{~m} / \mathrm{s}\) in the \(+x\) direction. So, the velocity vector can be written as \(\overrightarrow{\boldsymbol{v}} = 6.80 \hat{\imath} \mathrm{~m} / \mathrm{s}\)
02

Find the induced emf

To calculate the induced emf, we can use Faraday's Law of electromagnetic induction or calculate the cross product of the velocity and magnetic field vectors. The latter approach will be applied in this step. The cross product of the magnetic field vector and the velocity vector is given by \(\overrightarrow{\boldsymbol{E}} = \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}\), where \(\overrightarrow{\boldsymbol{E}}\) is the induced emf, \(\overrightarrow{\boldsymbol{v}}\) is velocity vector and \(\overrightarrow{\boldsymbol{B}}\) is magnetic field vector. Hence we get, \(\overrightarrow{\boldsymbol{E}} = \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}} = 6.80 \hat{\imath} \times ((0.120 \hat{\imath} - 0.220 \hat{\jmath} - 0.090 \hat{\boldsymbol{k}}) = 0.612 \hat{\jmath}+0.306 \hat{\boldsymbol{k}} \mathrm{V/m}\). Now, to find the total potential difference or induced emf we need to multiply \(\overrightarrow{\boldsymbol{E}}\) with the length of the rod (\(L = 25.0 \mathrm{cm} = 25.0 \times 10^{-2} \mathrm{m}\)). Hence, induced emf \( = E \times L = \sqrt{ (0.612)^2 + (0.306)^2} \times 25.0 \times 10^{-2} = 0.120 \mathrm{mV}\)
03

Direction of induced emf

Using right hand rule, the direction of the induced emf (\(\Delta V= V_B - V_A\)) will be from point B to point A where B is at higher potential and A is at lower potential. Now looking at the given axes, as the rod moves, the lower end (near the origin of the axes) will be at a higher potential while the upper end (away from the origin of the axes) will be at lower potential.

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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 of electromagnetic induction is a fundamental principle that describes how electric currents are induced in conductors due to changing magnetic fields. When a conductor, like a metal rod, is situated within a magnetic field and there is relative motion between them, an electromotive force (emf) is generated within the conductor. The strength of this induced emf is directly proportional to the rate of change of magnetic flux through the conductor.

Mathematically, Faraday's Law is expressed as follows:
\[\begin{equation} \varepsilon = - \frac{d\Phi_B}{dt} \end{equation}\]
Here, \(\varepsilon\) represents the induced emf, and \(d\Phi_B/dt\) is the rate of change of magnetic flux \(\Phi_B\) over time \(t\). The negative sign indicates the direction of the induced emf is such that it opposes the change in flux, according to Lenz's Law. In the classroom exercise, the magnetic field is uniform and constant, so the change in flux comes from the rod's motion through the field.
Induced Emf
The term 'induced emf' refers to the voltage generated across a conductor when it cuts through magnetic field lines. In the provided exercise, the metal rod moves through a magnetic field, leading to a separation of charges within the rod due to the magnetic forces acting on the moving charges. This separation of charges creates an electrical potential difference, known as induced emf. The induced emf can be calculated without directly using Faraday's Law by applying the formula that involves the cross product of velocity and magnetic field vectors:

\[\begin{equation} \varepsilon = \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}} \cdot \overrightarrow{L} \end{equation}\]
where \(\overrightarrow{\mathbf{v}}\) is the velocity vector of the conductor, \(\overrightarrow{\mathbf{B}}\) is the magnetic field vector, and \(\overrightarrow{L}\) is the length vector of the conductor. In simpler terms, induced emf is the voltage created when a conductor moves in a manner that intersects the magnetic field lines.
Magnetic Field
A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. Often visualized by field lines, a magnetic field is mathematically defined by the vector \(\overrightarrow{\mathbf{B}}\), and its strength and direction can vary throughout space. Magnetic fields can be produced by permanent magnets or by electric currents passing through a conductor, which underlie many electrical and electronic devices.

In the given exercise, the uniform magnetic field in which the rod is moving is characterized by both magnitude and direction, given by the vector \(\overrightarrow{\mathbf{B}} = (0.120 \ \mathrm{T}) \hat{\imath} - (0.220 \ \mathrm{T}) \hat{\jmath} - (0.0900 \ \mathrm{T}) \hat{\mathbf{k}}\). The interaction of this field with the rod's motion results in electromagnetic induction.
Cross Product
The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space, producing a third vector that is perpendicular to the plane of the input vectors. Importantly, the magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors.

The cross product is algebraically defined as:\[\begin{equation} \overrightarrow{\mathbf{C}} = \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}} \end{equation}\]
Here, \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) are vectors, and \(\overrightarrow{\mathbf{C}}\) is their cross product. If \(\overrightarrow{\mathbf{A}} = a_1\hat{\imath} + a_2\hat{\jmath} + a_3\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}} = b_1\hat{\imath} + b_2\hat{\jmath} + b_3\hat{\mathbf{k}}\), the cross product is given by:\[\begin{equation} \overrightarrow{\mathbf{C}} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{\mathbf{k}} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \end{equation}\]
In the context of our exercise, the cross product is vital for calculating the induced emf, as the velocity of the rod is crossed with the magnetic field vector to find the electric field vector, which is then used to determine the emf across the length of the rod.

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Most popular questions from this chapter

A copper wire has a square cross \(2.3 \mathrm{~mm}\) on a side. The wire is \(4.0 \mathrm{~m}\) long and carries a current of \(3.6 \mathrm{~A}\). The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

A closely wound rectangular coil of 80 turns has dimensions of \(25.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm} .\) The plane of the coil is rotated from a position where it makes an angle of \(37.0^{\circ}\) with a magnetic field of \(1.70 \mathrm{~T}\) to a position perpendicular to the field. The rotation takes \(0.0600 \mathrm{~s}\). What is the average emf induced in the coil?

A battery has emf \(30.0 \mathrm{~V}\) and internal resistance \(r .\) A \(9.00 \Omega\) resistor is connected to the terminals of the battery, and the voltage drop across the resistor is \(27.0 \mathrm{~V}\). What is the internal resistance of the battery?

A 5.00 A current runs through a 12 gauge copper wire (diameter \(2.05 \mathrm{~mm}\) ) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Light Bulbs. The power rating of a light bulb (such as a \(100 \mathrm{~W}\) bulb is the power it dissipates when connected across a \(120 \mathrm{~V}\) potential difference. What is the resistance of (a) a \(100 \mathrm{~W}\) bulb and (b) a \(60 \mathrm{~W}\) bulb? (c) How much current does each bulb draw in normal use?

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