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).

Step by step solution

01

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: \[V_{ind} = B\ell v\]

02

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: \[i_{ind} = \frac{B\ell v}{R}\]

03

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: \[F_{ext} = i_{ind} \ell B = \frac{B^2\ell^2 v}{R}\]

04

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: \[P_{ext} = \frac{B^2\ell^2 v^2}{R}\]

05

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.

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