Question

Two parallel conducting rails with negligible resistance are connected at one end by a resistor of resistance \(R\), as shown in the figure. The rails are placed in a magnetic field \(\vec{B}_{\text {ext }},\) which is perpendicular to the plane of the rails. This magnetic field is uniform and time independent. The distance between the rails is \(\ell\). A conducting rod slides without friction on top of the two rails at constant velocity \(\vec{v}\). a) Using Faraday's Law of Induction, calculate the magnitude of the potential difference induced in the moving rod. b) Calculate the magnitude of the induced current in the \(\operatorname{rod}, i_{\text {ind }}\). c) Show that for the rod to move at a constant velocity as shown, it must be pulled with an external force, \(\vec{F}_{\mathrm{ext}},\) and calculate the magnitude of this force. d) Calculate the work done, \(W_{\text {ext }},\) and the power generated, \(P_{\text {ext }}\), by the external force in moving the rod. e) Calculate the power used (dissipated) by the resistor, \(P_{\mathrm{R}}\). Explain the correlation between this result and those of part (d).

Short answer

Question: Calculate the power dissipated by the resistor when a rod of length (\(\ell\)) moves at a constant velocity (\(v\)) through a uniform magnetic field (\(B\)) along two parallel horizontal rails connected by a resistor (\(R\)). Answer: The power dissipated by the resistor (\(P_R\)) can be calculated using the formula: \[P_R =\frac{B^2\ell^2 v^2}{R}\]

Step-by-step solution

Find the potential difference induced in the moving rod

Use Faraday's Law of Induction, which states that the induced electromotive force (EMF) in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In this case, the magnetic flux changes as the rod moves across the rails. The magnetic flux (\(\Phi\)) through the loop formed by the rod and rails can be expressed as: \(\Phi = B \cdot A\), where B is the magnetic field and A is the area enclosed by the loop. Since the magnetic field is uniform and perpendicular to the rails, and the area enclosed by the loop is \(\ell \times x\), where x is the displacement of the rod from the left rail, the magnetic flux can be written as \(\Phi = B \cdot\ell\cdot x\). Now, find the time derivative of the flux to find the induced EMF (we drop the negative sign because we are looking for the magnitude of the potential difference): \[\frac{d\Phi}{dt} = B\ell\frac{dx}{dt}\] Since \(\frac{dx}{dt}\) is equal to the constant velocity \(v\), the magnitude of the potential difference (induced EMF) is: \[V_{ind} = B\ell v\]

Calculate the induced current

Use Ohm's Law to find the induced current in the rod: \(V_{ind} = i_{ind}R\) Now we can substitute the expression for the potential difference from Step 1 to calculate the induced current: \[i_{ind} = \frac{B\ell v}{R}\]

Find the external force

The motion of the rod through the magnetic field generates a Lorentz force, \(\vec{F}_{B}\), acting on the charges and producing a current. According to Newton's Second Law, this force must be balanced by an external force, \(\vec{F}_{ext}\), for the rod to move at a constant velocity. The Lorentz force can be calculated as: \(\vec{F}_{B} = I\vec{l} \times \vec{B}\), where I is the induced current and \(\vec{l}\) is a vector with length equal to the length of the rod and direction perpendicular to the rails. Since the rod and rails are parallel and current is in the same direction as \(\vec{l}\), the Lorentz force \(F_{B}\) is perpendicular to both \(\vec{l}\) and \(\vec{B}\), and its magnitude can be expressed as: \(F_{B} = i_{ind} \ell B\) To maintain a constant velocity, the magnitude of the external force, \(F_{ext}\), must be equal to the Lorentz force: \(F_{ext} = F_{B}\) Therefore, the magnitude of the external force is: \[F_{ext} = i_{ind} \ell B = \frac{B^2\ell^2 v}{R}\]

Calculate the work done and power generated by the external force

The work done by the external force \(W_{ext}\) can be calculated as \(W_{ext} = F_{ext} \cdot x\). Since the force is constant during the motion, we can write: \[W_{ext} = \frac{B^2\ell^2 v}{R} \cdot x\] The power generated by the external force, \(P_{ext}\), is the rate of work done: \[P_{ext} = \frac{dW_{ext}}{dt} = \frac{B^2\ell^2 v}{R}\frac{dx}{dt}\] Since \(\frac{dx}{dt} = v\), the power generated by the external force is: \[P_{ext} = \frac{B^2\ell^2 v^2}{R}\]

Calculate the power dissipated by the resistor and find the correlation

The power dissipated by the resistor, \(P_R\), can be expressed as \(P_R = i_{ind}^2 R\). Substituting the induced current from Step 2, we get: \[P_R = (\frac{B\ell v}{R})^2 R = \frac{B^2\ell^2 v^2}{R}\] Comparing the results from Steps 4 and 5, we can see that the power dissipated by the resistor (\(P_R\)) is equal to the power generated by the external force (\(P_{ext}\)). This implies that all the work done by the external force is used to generate energy that is ultimately dissipated by the resistor as heat.

Key concepts

Faraday's Law of Induction

When we talk about Faraday's Law of Induction, we're diving into how electric currents can be generated by changing magnetic fields. This law is fundamental in electromagnetic induction and is actually one of Maxwell's equations, which are essential to understand the physics of electricity and magnetism.

Faraday's Law of Induction states that the electromotive force (EMF) induced in a closed loop is proportional to the rate of change of magnetic flux through the loop. The formula for this law is given by:\[ \varepsilon = - \frac{d\Phi}{dt} \]where \(\varepsilon\) is the induced EMF and \(\Phi\) is the magnetic flux. In simpler terms, if we have a coil and a magnet is moved through it, a voltage is generated due to the change in magnetic field. In our example of the sliding rod, as it moves over the parallel rails, the area of the loop formed by the rod and rails changes, resulting in a changing magnetic flux. This is why an EMF is induced in the rod.

This principle is the cornerstone for the operation of generators, transformers, and many other electrical devices. Understanding it not only helps with solving textbook problems, but also provides insight into real-world applications.

Lorentz Force

The Lorentz Force is a crucial concept when dealing with charged particles in electromagnetic fields. This force is essentially what causes charged particles to move in response to electromagnetic fields. The Lorentz Force can be described by the formula:\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \]where \(q\) is the charge of the particle, \(\vec{E}\) is the electric field, \(\vec{v}\) is the velocity of the charged particle, and \(\vec{B}\) is the magnetic field.

In the context of the sliding rod problem, there is no electric field, but the movement of the rod through the magnetic field induces a current. This current carrying conductor in a magnetic field experiences the Lorentz Force, which is perpendicular to both the velocity of the rod and the magnetic field.

The magnitude of this force depends on the current, the length of the conductor within the magnetic field, and the strength of the field itself. For the rod to maintain a constant velocity, an external force must counteract this Lorentz Force, thereby explaining the necessity of \(\vec{F}_{\text{ext}}\). This is a clear demonstration of how magnetic fields can exert forces and influence the motion of conductive objects.

Ohm's Law

Ohm's Law is one of the simplest yet most important rules of electronics and physics when it comes to understanding how electric circuits operate. It establishes a relationship between voltage, current, and resistance in a conductor. Mathematically, it is expressed as:\[ V = IR \]where \(V\) is the voltage across the conductor, \(I\) is the current flowing through it, and \(R\) is its resistance.

In our textbook example, the induced EMF generated as the rod slides across the rails causes a current to flow through the circuit formed by the rails and resistor. We use Ohm's Law to find this induced current by rearranging the formula:\[ i_{\text{ind}} = \frac{V_{\text{ind}}}{R} \]This equation highlights that the current is directly proportional to the induced voltage and inversely proportional to the resistance.

Understanding Ohm’s Law is pivotal when analyzing series and parallel circuits, as it helps predict how circuits will behave when variables like voltage and resistance are changed. This principle is foundational for any troubleshooting or design work involving electrical circuits.

Power Dissipation

In electrical systems, power dissipation is an important concept, as it represents how electrical energy is converted into heat within a circuit. Knowing how to calculate and manage power dissipation is vital for ensuring that electronic devices operate safely and efficiently.

Power dissipated in a resistor, for instance, can be found using the formula:\[ P = I^2 R \]In our circuit with the sliding rod, the current established by the induced EMF causes power dissipation primarily as heat in the resistor. This dissipation follows the energy conservation principle, showing how the energy supplied by the external force (to counteract the Lorentz Force) is converted entirely into heat.

By analyzing the given resistor power formula, it's clear that this result is consistent with the power provided by the external force. The exercise shows a classic example of energy transformation from mechanical work (through the exerted external force on the rod) into electrical energy (current) and finally into thermal energy. Understanding power dissipation not only explains energy loss in circuits but also aids in designing components that can handle specific power levels without damage.