Difference between revisions of "E & M Qual Problems"

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1.) A rigid, infinitesimally thin conducting ring of radius <math>a=0.15 m</math> and mass <math>m=1.0kg</math> is initially oriented nearly vertically in a uniform vertical magnetic field <math>\vec{B} = B_0 \hat{z}</math> where <math>B_o = 2.5 Telsa</math>.  The resistance of the ring is <math>R = 10^{-3} \Omega</math>
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1.) A rigid, infinitesimally thin conducting ring of radius <math>a=0.15 m</math> and mass <math>m=1.0kg</math> is initially oriented nearly vertically in a uniform vertical magnetic field <math>\vec{B} = B_0 \hat{z}</math> where <math>B_o = 2.5 Telsa</math>.  The resistance of the ring is <math>R = 10^{-3} \Omega</math>.  The ring is initially at rest at a polar angle of <math>\theta_0 = 1.0 mrad</math> with respect to the <math>\hat{z}</math>-axis.  The orientation of the rin is such that the unit normal vector <math>\hat{n}</math> lies in the <math>\hat{y} -\hat{z}</math> plane, initially nearly parallel to the <math>-\hat{y}</math> direction.
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If the ring is allowed to fall with no slipping under the influence of the earth's gravitational field, the presence of the magnetic field acts as a magnetic "brake", slowing the fall of the ring by a considerable amount, as if the ring was falling in a highly viscous fluid.
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a.) Calculate the instantaneous EMP <math>\epsilon(t)</math> induced in the ring in terms of the polar angle <math>\theta</math> and the angular velocity <math>\dot{\theta}</math>.
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b.) Calculate the instantaneous current <math>I(t)</math> flowing inthe ring due to the induced EMF.  Express your answer in terms of <math>\theta</math> and  <math>\dot{\theta}</math>.
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c.) Derive the equation of motion for the ring by consideration of the torques acting on the ring.  Note that the moment of inertia for the ring, rotating about its side is <math>I=\frac{3}{2} ma^2</math>

Latest revision as of 20:58, 22 August 2007

1.) A rigid, infinitesimally thin conducting ring of radius [math]a=0.15 m[/math] and mass [math]m=1.0kg[/math] is initially oriented nearly vertically in a uniform vertical magnetic field [math]\vec{B} = B_0 \hat{z}[/math] where [math]B_o = 2.5 Telsa[/math]. The resistance of the ring is [math]R = 10^{-3} \Omega[/math]. The ring is initially at rest at a polar angle of [math]\theta_0 = 1.0 mrad[/math] with respect to the [math]\hat{z}[/math]-axis. The orientation of the rin is such that the unit normal vector [math]\hat{n}[/math] lies in the [math]\hat{y} -\hat{z}[/math] plane, initially nearly parallel to the [math]-\hat{y}[/math] direction.

If the ring is allowed to fall with no slipping under the influence of the earth's gravitational field, the presence of the magnetic field acts as a magnetic "brake", slowing the fall of the ring by a considerable amount, as if the ring was falling in a highly viscous fluid.

a.) Calculate the instantaneous EMP [math]\epsilon(t)[/math] induced in the ring in terms of the polar angle [math]\theta[/math] and the angular velocity [math]\dot{\theta}[/math].

b.) Calculate the instantaneous current [math]I(t)[/math] flowing inthe ring due to the induced EMF. Express your answer in terms of [math]\theta[/math] and [math]\dot{\theta}[/math].

c.) Derive the equation of motion for the ring by consideration of the torques acting on the ring. Note that the moment of inertia for the ring, rotating about its side is [math]I=\frac{3}{2} ma^2[/math]