218 lines
13 KiB
TeX
218 lines
13 KiB
TeX
\subsection{Force on Current-Carrying Conductors and Loops}
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A current-carrying wire in a magnetic field experiences a force because the moving charge carriers inside the wire each feel a magnetic force. When the current flows through a closed loop, the forces on individual segments can produce a net torque, causing the loop to rotate. This is the operating principle of electric motors.
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\dfn{Force on a current-carrying wire}{A wire carrying current $I$ and placed in a magnetic field $\vec{B}$ experiences a magnetic force. For a straight wire segment of length $\ell$ carrying current $I$, with the vector $\vec{\ell}$ pointing in the direction of the current, the force is
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\[
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\vec{F} = I\,\vec{\ell}\times\vec{B}.
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\]
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The magnitude of this force is
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\[
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F = I\,\ell\,B\,\sin\theta,
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\]
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where $\theta$ is the angle between the current direction (the direction of $\vec{\ell}$) and the magnetic field $\vec{B}$. The direction is given by the right-hand rule for cross products, reversed for negative current carriers.}
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\nt{The vector $\vec{\ell}$ has magnitude equal to the length of the wire segment and points along the wire in the direction of conventional current. In a curved wire, the total force is obtained by integrating the differential-force expression over the entire path: $\vec{F} = \displaystyle\oint I\,d\vec{\ell}\times\vec{B}$.}
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\thm{Magnetic force on a current-carrying conductor}{Let a wire carry steady current $I$ through a magnetic field $\vec{B}$.
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\begin{itemize}
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\item \textbf{Straight wire:} For a straight wire segment of length $\ell$, with $\vec{\ell}$ pointing along the wire in the current direction,
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\[
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\vec{F} = I\,\vec{\ell}\times\vec{B}.
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\]
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The magnitude is $F = I\,\ell\,B\,\sin\theta$, where $\theta$ is the angle between $\vec{\ell}$ and $\vec{B}$.
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\item \textbf{Differential element:} For an arbitrary wire path, the force on a differential element $d\vec{\ell}$ is
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\[
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d\vec{F} = I\,d\vec{\ell}\times\vec{B},
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\]
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and the total force is
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\[
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\vec{F} = \int_{\text{wire}} I\,d\vec{\ell}\times\vec{B}.
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\]
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\item \textbf{Right-hand rule:} Point your right hand's fingers along $\vec{\ell}$ (current direction), then curl them toward $\vec{B}$. Your thumb gives the direction of $\vec{F}$.
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\end{itemize}}
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\pf{Force on a current-carrying wire from the force on moving charges}{The force on a single charge $q$ moving with drift velocity $\vec{v}_d$ is $\vec{F}_q = q\,\vec{v}_d\times\vec{B}$. In a wire segment of length $\ell$ and cross-sectional area $A$, the number of charge carriers is $N = n\,A\,\ell$, where $n$ is the carrier number density. The total force is
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\[
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\vec{F} = N\,q\,\vec{v}_d\times\vec{B} = n\,A\,\ell\,q\,\vec{v}_d\times\vec{B}.
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\]
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The current is $I = n\,q\,v_d\,A$, and the direction of $\vec{v}_d$ for positive carriers is the current direction. Letting $\vec{\ell}$ point in that direction with magnitude $\ell$, we have $n\,q\,\vec{v}_d = (I/A)\,\hat{\ell}$, and so
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\[
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\vec{F} = I\,\vec{\ell}\times\vec{B},
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\]
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where we used $\vec{\ell} = \ell\,\hat{\ell}$. This derivation confirms that the macroscopic force on a current-carrying wire follows directly from the Lorentz force on individual charge carriers.}
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\cor{Straight wire parallel or perpendicular to field}{When the wire is parallel or antiparallel to $\vec{B}$ ($\theta = 0^\circ$ or $180^\circ$), then $\sin\theta = 0$ and $F = 0$. When the wire is perpendicular to $\vec{B}$ ($\theta = 90^\circ$), the force is maximal: $F = I\,\ell\,B$.}
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\cor{Closed loop in a uniform field}{When a closed current loop sits entirely in a uniform magnetic field, the net force is zero:
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\[
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\vec{F}_{\text{net}} = I\oint d\vec{\ell}\times\vec{B} = I\left(\oint d\vec{\ell}\right)\times\vec{B} = \vec{0}.
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\]
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The integral $\oint d\vec{\ell}$ around any closed loop is the zero vector. Thus, no net force acts on a closed loop in a uniform field, though individual segments still feel forces.}
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\dfn{Magnetic dipole moment of a current loop}{A planar loop carrying current $I$ with enclosed area $A$ has a \emph{magnetic dipole moment}
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\[
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\vec{\mu} = N\,I\,A\,\hat{n},
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\]
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where $N$ is the number of turns in the loop, $A$ is the area enclosed by one turn, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop. The direction of $\hat{n}$ is given by the right-hand rule: curl the fingers of your right hand around the loop in the direction of the current, and your thumb points along $\hat{n}$.}
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\mprop{Torque and potential energy of a current loop in a uniform field}{Let a planar current loop with magnetic dipole moment $\vec{\mu} = N I A\,\hat{n}$ be placed in a uniform magnetic field $\vec{B}$. Then:
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\begin{enumerate}
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\item The torque on the loop is
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\[
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\vec{\tau} = \vec{\mu}\times\vec{B}.
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\]
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The magnitude is
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\[
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\tau = \mu\,B\,\sin\phi = N\,I\,A\,B\,\sin\phi,
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\]
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where $\phi$ is the angle between $\hat{n}$ and $\vec{B}$.
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\item The potential energy of the loop is
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\[
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U = -\vec{\mu}\cdot\vec{B} = -\mu\,B\,\cos\phi = -N\,I\,A\,B\,\cos\phi.
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\]
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\end{enumerate}
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Equilibrium occurs when $\phi = 0^\circ$ (stable, $\hat{n}\parallel\vec{B}$, torque zero, energy minimum) or $\phi = 180^\circ$ (unstable, $\hat{n}$ antiparallel to $\vec{B}$, torque zero, energy maximum).}
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\pf{Torque on a current loop in a uniform field}{Consider a rectangular loop of width $a$ (in the $x$-direction) and height $b$ (in the $y$-direction), carrying current $I$ in a uniform field $\vec{B}=B\,\hat{k}$ (out of the plane). Let the normal $\hat{n}$ to the loop make angle $\phi$ with $\hat{k}$. The loop rotates about an axis through its center, perpendicular to $\vec{B}$.
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The forces on the top and bottom segments (length $a$) are equal and opposite and collinear, so they cancel. The forces on the two side segments (length $b$) are
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\[
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\vec{F}_1 = I\,\vec{b}_1\times\vec{B}
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\quad\text{and}\quad
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\vec{F}_2 = I\,\vec{b}_2\times\vec{B},
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\]
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with $\vec{b}_1 = -\vec{b}_2$. These forces have magnitude $F = I\,b\,B$ and act at perpendicular distances $(a/2)\,\sin\phi$ from the axis. Each produces a torque of magnitude
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\[
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\tau_{\text{side}} = F\cdot\frac{a}{2}\,\sin\phi = I\,b\,B\cdot\frac{a}{2}\,\sin\phi.
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\]
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Both sides contribute in the same rotational sense, so
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\[
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\tau = 2\cdot I\,b\,B\cdot\frac{a}{2}\,\sin\phi = I\,a\,b\,B\,\sin\phi.
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\]
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Since $A = a\,b$ and $\vec{\mu}$ has magnitude $\mu = I A$ (for $N=1$),
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\[
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\tau = \mu\,B\,\sin\phi.
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\]
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The vector form is $\vec{\tau} = \vec{\mu}\times\vec{B}$. For $N$ turns, multiply by $N$.
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For the potential energy, the torque tends to align $\hat{n}$ with $\vec{B}$. The work done by an external agent rotating the loop from angle $\phi_0$ to $\phi$ equals the change in potential energy:
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\[
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\Delta U = -\int_{\phi_0}^{\phi} \tau_{\text{ext}}\,d\phi'
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= -\int_{\phi_0}^{\phi} \mu B\,\sin\phi'\,d\phi'
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= -\mu B\left(\cos\phi_0 - \cos\phi\right).
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\]
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Choosing the reference $U(\phi_0 = 90^\circ) = 0$, we find
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\[
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U = -\mu B\,\cos\phi = -\vec{\mu}\cdot\vec{B}.
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\]
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\nt{The torque tries to align $\vec{\mu}$ with $\vec{B}$. The stable equilibrium orientation has $\hat{n}\parallel\vec{B}$, i.e., the plane of the loop is perpendicular to the field. The unstable equilibrium has $\hat{n}$ antiparallel to $\vec{B}$. A DC motor exploits this by periodically reversing the current so the loop continues rotating.}}
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\nt{The SI unit of magnetic dipole moment $\vec{\mu}$ is ampere-square meter, $\mathrm{A\cdot m^2}$. This is equivalent to joules per tesla, $\mathrm{J/T}$.}
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\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:
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\[
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\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.}
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\qs{Worked example}{A rectangular wire loop has width $a=0.10\,\mathrm{m}$ (horizontal side) and height $b=0.050\,\mathrm{m}$ (vertical side). The loop has $N=100$ turns and carries current $I=2.0\,\mathrm{A}$ in the direction shown: clockwise when viewed from the front (from the $+x$-axis toward the $yz$-plane). The loop sits in a uniform magnetic field $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$ pointing to the right. The normal vector $\hat{n}$ points along the $-\hat{k}$ direction (into the page) by the right-hand rule for clockwise current.
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The loop lies in the $xy$-plane, centered at the origin, with its sides parallel to the $x$- and $y$-axes.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the net magnetic force on the loop,
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\item the net magnetic torque on the loop (magnitude and direction), and
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\item the magnetic potential energy of the loop in this orientation, and determine whether the loop will rotate clockwise or counterclockwise when released.
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\end{enumerate}}
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\textbf{Given quantities:}
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\begin{itemize}
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\item Width (horizontal): $a = 0.10\,\mathrm{m}$
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\item Height (vertical): $b = 0.050\,\mathrm{m}$
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\item Number of turns: $N = 100$
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\item Current: $I = 2.0\,\mathrm{A}$ (clockwise in the $xy$-plane)
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\item Magnetic field: $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$
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\item Normal vector: $\hat{n} = -\hat{k}$
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\end{itemize}
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\sol \textbf{(a) Net force.} The magnetic field is uniform, and the loop is a closed current loop. From the corollary, the net force on a closed loop in a uniform magnetic field is zero. We can verify this segment by segment.
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The loop has four segments:
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\begin{itemize}
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\item \textit{Bottom segment} (length $a$, current to the right): $\vec{\ell}_1 = a\,\hat{\imath}$, so $\vec{F}_1 = I\,(a\,\hat{\imath})\times(B\,\hat{\imath}) = \vec{0}$ (parallel to $\vec{B}$).
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\item \textit{Top segment} (length $a$, current to the left): $\vec{\ell}_3 = -a\,\hat{\imath}$, so $\vec{F}_3 = I\,(-a\,\hat{\imath})\times(B\,\hat{\imath}) = \vec{0}$.
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\item \textit{Right segment} (length $b$, current downward): $\vec{\ell}_2 = -b\,\hat{\jmath}$, so
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\[
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\vec{F}_2 = I\,(-b\,\hat{\jmath})\times(B\,\hat{\imath}) = I\,b\,B\,(-\hat{\jmath}\times\hat{\imath}) = I\,b\,B\,\hat{k}.
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\]
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\item \textit{Left segment} (length $b$, current upward): $\vec{\ell}_4 = b\,\hat{\jmath}$, so
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\[
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\vec{F}_4 = I\,(b\,\hat{\jmath})\times(B\,\hat{\imath}) = I\,b\,B\,(\hat{\jmath}\times\hat{\imath}) = -I\,b\,B\,\hat{k}.
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\]
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\end{itemize}
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Summing:
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\[
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\vec{F}_{\text{net}} = \vec{F}_1+\vec{F}_2+\vec{F}_3+\vec{F}_4 = \vec{0} + I\,b\,B\,\hat{k} + \vec{0} - I\,b\,B\,\hat{k} = \vec{0}.
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\]
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For $N$ turns, each segment's force is multiplied by $N$, but they still cancel:
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\[
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\boxed{\vec{F}_{\text{net}} = \vec{0}}.
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\]
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\textbf{(b) Net torque.} The magnetic dipole moment has magnitude
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\[
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\mu = N\,I\,A = N\,I\,(a\,b).
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\]
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Substitute the values:
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\[
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\mu = (100)(2.0\,\mathrm{A})(0.10\,\mathrm{m})(0.050\,\mathrm{m}) = (100)(2.0)(0.0050)\,\mathrm{A\cdot m^2} = 1.0\,\mathrm{A\cdot m^2}.
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\]
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The dipole moment vector is $\vec{\mu} = \mu\,\hat{n} = -(1.0\,\mathrm{A\cdot m^2})\,\hat{k}$.
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The torque is
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\[
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\vec{\tau} = \vec{\mu}\times\vec{B} = [-(1.0)\,\hat{k}]\times[(0.60)\,\hat{\imath}] = -(1.0)(0.60)\,(\hat{k}\times\hat{\imath}).
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\]
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Since $\hat{k}\times\hat{\imath}=\hat{\jmath}$,
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\[
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\vec{\tau} = -(0.60)\,\hat{\jmath}\,\mathrm{N\cdot m}.
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\]
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The magnitude is
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\[
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\tau = 0.60\,\mathrm{N\cdot m}.
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\]
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To interpret the direction, $-\hat{\jmath}$ points downward. Using the right-hand rule for torque, the loop tends to rotate such that $\vec{\mu}$ aligns with $\vec{B}$ — that is, $\hat{n}$ rotates from $-\hat{k}$ toward $+\hat{\imath}$. This corresponds to a rotation about the $y$-axis.
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More physically: the right side of the loop (at $+x/2$) feels force $\vec{F}_2 = N\,I\,b\,B\,\hat{k} = 100(2.0)(0.050)(0.60)\,\hat{k} = 6.0\,\mathrm{N}$ upward, while the left side (at $-x/2$) feels $6.0\,\mathrm{N}$ downward. This pair of forces creates a torque that rotates the right side up and the left side down — a rotation about the $y$-axis.
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\[
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\boxed{\tau = 0.60\,\mathrm{N\cdot m},\quad \text{rotation about the }-\hat{\jmath}\text{ axis (right side up, left side down)}}.
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\]
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\textbf{(c) Potential energy and rotational tendency.} The potential energy is
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\[
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U = -\vec{\mu}\cdot\vec{B} = -[-(1.0)\,\hat{k}]\cdot[(0.60)\,\hat{\imath}].
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\]
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Since $\hat{k}\perp\hat{\imath}$, their dot product is $0$, so
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\[
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\boxed{U = 0}.
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\]
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This corresponds to $\phi = 90^\circ$ between $\vec{\mu}$ (pointing in $-\hat{k}$) and $\vec{B}$ (pointing in $+\hat{\imath}$), since $\cos 90^\circ = 0$.
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The loop will rotate toward the stable equilibrium orientation where $\vec{\mu}\parallel\vec{B}$. Currently $\vec{\mu}$ points into the page ($-\hat{k}$) while $\vec{B}$ points right ($+\hat{\imath}$). The stable orientation has $\hat{n}$ aligned with $+\hat{\imath}$, meaning the loop's plane is perpendicular to the field (normal pointing along $\vec{B}$). The torque computed in part~(b) drives this rotation: the right side is pushed upward and the left side downward, rotating the loop about the $y$-axis. Viewed from the $+y$ direction (looking down from above), this rotation appears counterclockwise.
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\[
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\boxed{\text{Rotates about the }y\text{-axis toward stable equilibrium (right side up, left side down).}}
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\]
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\bigskip
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\textbf{Summary of results:}
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\begin{enumerate}[label=(\alph*)]
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\item $\vec{F}_{\text{net}} = \vec{0}$
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\item $\tau = 0.60\,\mathrm{N\cdot m}$, rotation about the $-\hat{\jmath}$ axis
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\item $U = 0$, loop rotates toward stable equilibrium where $\hat{n}\parallel\vec{B}$
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\end{enumerate}
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