206 lines
11 KiB
TeX
206 lines
11 KiB
TeX
\subsection{Motional Electromotive Force}
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When a conductor moves through a magnetic field, the magnetic force on the charge carriers inside the conductor separates charge and produces an electromotive force. This effect, called \emph{motional emf}, is one of the two fundamental mechanisms of electromagnetic induction (the other being a time-varying magnetic field).
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\dfn{Motional emf}{When a conductor moves with velocity $\vec{v}$ through a magnetic field $\vec{B}$, each charge carrier of charge $q$ experiences the magnetic force
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\[
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\vec{F}_B = q\,\vec{v}\times\vec{B}.
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\]
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This force acts as a non-electrostatic force per unit charge, $\vec{v}\times\vec{B}$, which drives charges along the conductor. The \emph{motional emf} along a path from point $a$ to point $b$ within the conductor is the line integral
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\[
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\mathcal{E} = \int_{a}^{b} (\vec{v}\times\vec{B})\cdot d\vec{\ell},
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\]
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where $d\vec{\ell}$ is an infinitesimal displacement vector along the conductor from $a$ to $b$. The motional emf represents the work done per unit charge by the magnetic force as charges move along the conductor.}
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\nt{Even though the magnetic force itself does no net work on a moving charge (since $\vec{F}_B\perp\vec{v}$), the motional emf arises because the conductor's motion provides the energy transfer mechanism. The external agent pushing the conductor does mechanical work; the magnetic field acts as the intermediary that converts this work into electrical energy.}
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\nt{Inside the moving conductor (the ``source''), the emf drives current from lower potential to higher potential, just as a battery drives current from its negative to its positive terminal. The moving conductor thus acts as a battery with emf $\mathcal{E}$ and zero internal resistance (if the conductor is ideal).}
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\thm{Motional emf of a straight conductor in a uniform field}{Let a straight conductor of length $\ell$ move with constant velocity $\vec{v}$ through a uniform magnetic field $\vec{B}$.
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\begin{itemize}
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\item \textbf{General case:} The motional emf between the two ends of the conductor is
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\[
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\mathcal{E} = \int_{a}^{b} (\vec{v}\times\vec{B})\cdot d\vec{\ell},
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\]
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where the integral is taken along the conductor from end $a$ to end $b$.
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\item \textbf{Mutually perpendicular case:} When $\vec{v}$, $\vec{B}$, and the conductor are mutually perpendicular, the motional emf simplifies to
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\[
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\mathcal{E} = B\,\ell\,v.
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\]
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\item \textbf{Direction of emf:} The end of the conductor toward which positive charges are pushed by $\vec{v}\times\vec{B}$ is at higher potential. Use the right-hand rule for the cross product $\vec{v}\times\vec{B}$: point your fingers along $\vec{v}$, curl toward $\vec{B}$; your thumb points in the direction of $\vec{v}\times\vec{B}$, which is the direction positive charges move inside the conductor.
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\end{itemize}}
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\pf{Motional emf from the Lorentz force}{Consider a straight conducting bar of length $\ell$ moving with velocity $\vec{v}$ to the right through a uniform magnetic field $\vec{B}$ pointing out of the page. The bar is oriented vertically, perpendicular to both $\vec{v}$ and $\vec{B}$.
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Each conduction electron of charge $-e$ experiences a magnetic force
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\[
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\vec{F}_e = -e\,(\vec{v}\times\vec{B}).
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\]
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With $\vec{v}$ to the right ($+\hat{\imath}$) and $\vec{B}$ out of the page ($+\hat{k}$),
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\[
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\vec{v}\times\vec{B} = vB\,(\hat{\imath}\times\hat{k}) = -vB\,\hat{\jmath},
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\]
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so the force on electrons is $-e(-vB\,\hat{\jmath}) = evB\,\hat{\jmath}$, pointing \emph{upward} along the bar. Electrons accumulate at the top, leaving the bottom end positively charged.
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Charge separation continues until the resulting electrostatic field $\vec{E}_{\text{ind}}$ balances the magnetic force:
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\[
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-e\,\vec{E}_{\text{ind}} + (-e)(\vec{v}\times\vec{B}) = \vec{0}
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\quad\Rightarrow\quad
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\vec{E}_{\text{ind}} = -(\vec{v}\times\vec{B}).
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\]
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The motional emf is the line integral of the non-electrostatic force per unit charge along the bar. Equivalently, it equals the potential difference between the ends:
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\[
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\mathcal{E} = \int_{\text{bottom}}^{\text{top}} \vec{E}_{\text{ind}}\cdot d\vec{\ell} = E_{\text{ind}}\,\ell.
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\]
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Since $\vec{E}_{\text{ind}} = vB\,\hat{\jmath}$ and the bar extends along $\hat{\jmath}$, we obtain
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\[
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\mathcal{E} = vB\,\ell.
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\]
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More generally, without assuming perpendicularity,
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\[
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\mathcal{E} = \int_{a}^{b} (\vec{v}\times\vec{B})\cdot d\vec{\ell},
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\]
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which reduces to $B\ell v$ when $\vec{v}$, $\vec{B}$, and $d\vec{\ell}$ are mutually perpendicular. \Qed}
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\mprop{Motional emf in a circuit with resistance}{A conducting bar of length $\ell$ slides on frictionless conducting rails in a uniform magnetic field $B$, with the rails connected by a resistor $R$. The bar moves with instantaneous speed $v$ perpendicular to $B$. Then:
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\begin{enumerate}
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\item The motional emf is
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\[
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\mathcal{E} = B\,\ell\,v.
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\]
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\item The induced current in the circuit (magnitude) is
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\[
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I = \frac{\mathcal{E}}{R} = \frac{B\,\ell\,v}{R}.
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\]
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\item The magnetic force on the bar (magnitude) opposes the motion:
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\[
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F_{\text{mag}} = I\,\ell\,B = \frac{B^{2}\,\ell^{2}\,v}{R}.
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\]
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\item The external force needed to maintain constant velocity $v$ equals the magnetic force in magnitude:
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\[
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F_{\text{ext}} = \frac{B^{2}\,\ell^{2}\,v}{R}.
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\]
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\item Power dissipated in the resistor:
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\[
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P_{\text{elec}} = I^{2}R = \frac{B^{2}\,\ell^{2}\,v^{2}}{R}.
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\]
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\item Mechanical power delivered by the external agent:
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\[
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P_{\text{mech}} = F_{\text{ext}}\,v = \frac{B^{2}\,\ell^{2}\,v^{2}}{R}.
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\]
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\end{enumerate}
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Thus $P_{\text{mech}} = P_{\text{elec}}$, consistent with conservation of energy.}
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\cor{Motional emf and Faraday's law agree}{For a conducting bar sliding on rails, Faraday's law gives the induced emf as $\mathcal{E} = -\dfrac{d\Phi_B}{dt}$. If the bar moves a distance $dx$ in time $dt$, the area changes by $dA = \ell\,dx$ and the flux by $d\Phi_B = B\,\ell\,dx$, so
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\[
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\left\lvert\frac{d\Phi_B}{dt}\right\rvert = B\,\ell\,\frac{dx}{dt} = B\,\ell\,v.
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\]
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This is exactly the motional-emf result $\mathcal{E} = B\ell v$. Faraday's law provides the same emf through the ``flux-change'' perspective, while the motional-emf derivation provides it through the ``force-on-charges'' perspective.}
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\ex{Illustrative example}{A metal rod of length $\ell$ rotates with angular speed $\omega$ about one end in a uniform magnetic field $B$ perpendicular to the plane of rotation. Different points on the rod have different speeds $v(r) = r\,\omega$, so the emf must be computed by integration:
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\[
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\mathcal{E} = \int_{0}^{\ell} B\,(r\omega)\,dr = \frac{1}{2}\,B\,\omega\,\ell^{2}.\]}
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\qs{Worked example}{A conducting bar of length $\ell = 0.50\,\mathrm{m}$ slides on two horizontal, frictionless, conducting rails connected by a resistor $R = 4.0\,\Omega$ at the left end, as shown in the figure. A uniform magnetic field $\vec{B} = (0.30\,\mathrm{T})\,\hat{k}$ points straight upward (out of the horizontal plane of the rails). At a particular instant, the bar is moving to the right with velocity $v = 2.0\,\mathrm{m/s}$.
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\begin{enumerate}[label=(\alph*)]
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\item Find the motional emf induced in the circuit at this instant.
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\item Find the induced current: magnitude and direction (clockwise or counterclockwise as viewed from above).
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\item Find the magnitude and direction of the external force that must be applied to the bar to maintain this constant velocity.
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\item Find the power dissipated in the resistor. Verify that the mechanical power delivered by the external force equals the electrical power dissipated.
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\end{enumerate}}
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\textbf{Given quantities:}
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\begin{itemize}
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\item Bar length: $\ell = 0.50\,\mathrm{m}$
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\item Resistance: $R = 4.0\,\Omega$
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\item Magnetic field: $B = 0.30\,\mathrm{T}$ (out of page)
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\item Bar velocity: $v = 2.0\,\mathrm{m/s}$ (to the right)
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\item $\vec{v}$, $\vec{B}$, and the bar are mutually perpendicular.
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\end{itemize}
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\sol \textbf{(a) Motional emf.} Since $\vec{v}$, $\vec{B}$, and the bar are mutually perpendicular, the motional emf is
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\[
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\mathcal{E} = B\,\ell\,v.
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\]
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Substitute the given values:
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\[
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\mathcal{E} = (0.30\,\mathrm{T})(0.50\,\mathrm{m})(2.0\,\mathrm{m/s})
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= (0.30)(1.0)\,\mathrm{V} = 0.30\,\mathrm{V}.
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\]
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\textbf{(b) Induced current.} The magnitude of the induced current is given by Ohm's law:
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\[
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I = \frac{\mathcal{E}}{R} = \frac{0.30\,\mathrm{V}}{4.0\,\Omega}
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= 0.075\,\mathrm{A} = 75\,\mathrm{mA}.
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\]
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To find the direction, apply Lenz's law. The bar moves to the right, so the area of the loop increases, and the upward magnetic flux $\Phi_B$ through the loop is \emph{increasing}. Lenz's law requires the induced current to create a magnetic field $\vec{B}_{\text{ind}}$ that opposes this increase, so $\vec{B}_{\text{ind}}$ must point \emph{into} the page (downward). By the right-hand rule, a current that produces an into-the-page field flows \emph{clockwise} as viewed from above.
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Alternatively, use the force-on-charges argument: inside the moving bar, positive charge carriers experience $\vec{v}\times\vec{B}$, which points upward along the bar (since $\hat{\imath}\times\hat{k}=-\hat{\jmath}$ and positive charges move in the $+\hat{\jmath}$ direction if the bar is oriented in that direction). The upper end is thus at higher potential. Current flows from high to low potential through the external circuit (the resistor), i.e., clockwise.
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\[
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I = 0.075\,\mathrm{A},\quad \text{clockwise (as viewed from above)}.
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\]
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\textbf{(c) External force.} The induced current in the bar flows \emph{downward} (from the top rail to the bottom rail), i.e., in the $-\hat{\jmath}$ direction. Therefore, $\vec{\ell} = -\ell\,\hat{\jmath}$, and the magnetic force on the bar is
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\[
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\vec{F}_{\text{mag}} = I\,\vec{\ell}\times\vec{B} = I\,(-\ell\,\hat{\jmath})\times(B\,\hat{k})
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= -I\,\ell\,B\,(\hat{\jmath}\times\hat{k}) = -I\,\ell\,B\,\hat{\imath}.
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\]
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The magnetic force on the bar points to the \emph{left}, opposing the motion. Its magnitude is
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\[
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F_{\text{mag}} = I\,\ell\,B
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= (0.075\,\mathrm{A})(0.50\,\mathrm{m})(0.30\,\mathrm{T})
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= (0.075)(0.15)\,\mathrm{N} = 0.01125\,\mathrm{N}.
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\]
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To maintain constant velocity (zero net force), the external force must exactly balance the magnetic force:
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\[
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\vec{F}_{\text{ext}} = -\vec{F}_{\text{mag}} = +(0.01125\,\mathrm{N})\,\hat{\imath},
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\]
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i.e., $F_{\text{ext}} = 0.01125\,\mathrm{N}$ to the \emph{right}.
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Rounding to two significant figures (matching the given data):
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\[
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F_{\text{ext}} = 0.011\,\mathrm{N}\text{ to the right}.
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\]
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\textbf{(d) Power.} The electrical power dissipated in the resistor is
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\[
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P_{\text{elec}} = I^{2}R = (0.075\,\mathrm{A})^{2}(4.0\,\Omega)
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= (0.005625)(4.0)\,\mathrm{W} = 0.0225\,\mathrm{W}.
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\]
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Equivalently,
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\[
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P_{\text{elec}} = \frac{\mathcal{E}^{2}}{R}
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= \frac{(0.30\,\mathrm{V})^{2}}{4.0\,\Omega}
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= \frac{0.090}{4.0}\,\mathrm{W} = 0.0225\,\mathrm{W}.
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\]
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The mechanical power delivered by the external force is
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\[
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P_{\text{mech}} = F_{\text{ext}}\,v
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= (0.01125\,\mathrm{N})(2.0\,\mathrm{m/s})
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= 0.0225\,\mathrm{W}.
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\]
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Since $P_{\text{mech}} = P_{\text{elec}} = 0.0225\,\mathrm{W}$, mechanical energy is fully converted to electrical energy (Joule heating), as expected from energy conservation.
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\[
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P_{\text{elec}} = 0.0225\,\mathrm{W} = 22.5\,\mathrm{mW},
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\qquad
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P_{\text{mech}} = 0.0225\,\mathrm{W}.
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\]
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\bigskip
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\noindent\textbf{Summary of results:}
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\begin{enumerate}[label=(\alph*)]
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\item $\mathcal{E} = 0.30\,\mathrm{V}$
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\item $I = 0.075\,\mathrm{A} = 75\,\mathrm{mA}$, clockwise
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\item $F_{\text{ext}} = 0.011\,\mathrm{N}$ to the right
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\item $P_{\text{elec}} = 0.0225\,\mathrm{W}$, verified $P_{\text{mech}} = P_{\text{elec}}$
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\end{enumerate}
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