2001AugQualEM - Ben Sauerwine Practice for Qualifying Exams...

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Unformatted text preview: Ben Sauerwine Practice for Qualifying Exams Electricity and Magnetism Problem Source: CMU August 2001 Qualifying Exam (3) A solid glass sphere of mass M and radius R carries on its surface a uniformly distributed positive charge Q. It is spinning in outer space with angular speed ω . (a) Since the sphere is spinning, there are accelerated charges. Explain briefly why the spinning sphere does not emit electromagnetic radiation. Radiation is emitted when the acceleration has a component in the direction of velocity. The acceleration in the rotating sphere is perpendicular to the direction that the charges are moving in, and therefore no radiation is emitted in this case. (b) Next, an external magnetic field of magnitude B is established and the sphere moves in pure precession with the spin axis at an angle α to the magnetic field. Derive how much time it takes to precess through one complete turn. First I will find the magnetic moment of the sphere: ( ) ∫ × = x d x J x 3 2 1 v v v r μ Thus the magnetic moment of a loop with uniform current I is ( ) 2 ˆ 2 1 r I rd rI n π θ μ = = ∫ r Now integrating a stack of loops to obtain a sphere, I have: ( ) ( ) [ ] ( ) ( ) ( ) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = = = = = ∫ ∫ 8 3 4 ˆ sin 4 ˆ sin 4 sin sin ˆ 2 4 2 2 2 π ω φ φ ω μ φ ω π φ φ φ φ φ φ π φ μ π π R Q n d R Q n R R Q I R r d R r I n r r Now I may use this to determine the torque on the sphere: α ω π α μ μ τ sin 32 3 sin 2 B R Q B B = = × = r v the torque will be in the direction perpendicular to both the direction of rotation and the magnetic field. Now: ω τ v v v v s I L dt L d = = Note that since the torque is always perpendicular to the angular momentum, the angular momentum will precess about the magnetic field and never actually change. Finally, I must find the period of precession from this. In order to do so, I consider the total amount which the angular momentum must change in one precession: since ω s I L = , it will trace a circle of perimeter ( ) α ω π α π sin 2 sin 2 s I L = . Then, the time for one precession is s s I B QR I period 2 3 64 sin 2 π τ α ω π = = where ( ) ( ) 2 3 5 2 3 5 0 0 2 2 2 3 5 2 3 4 5 2 3 4 sin sin 5 2 3 4 sin sin 3 4 MR R MR d R MR dr d d r r R M I R s = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = ∫ ∫∫ ∫ π π φ φ φ π π φ θ φ φ π π π π QB M period π 15 128 = Note that this breaks down due to division by zero if = ω . Problem Source: CMU August 2001 Qualifying Exam (4) Alternating-Circuit Capacitor An ideal circular parallel-plate capacitor of radius a and plate separation d << a is connected to an alternating-current generator by axial leads, as shown in the sketch below. The current in the wire is ( ) ( ) t I t I ω sin = , where a c << ω and c is the speed of light. Assume that the axial leads lie along the z-axis with the origin at the center of the capacitor. (a) The electric and magnetic fields inside the capacitor between the plates have...
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2001AugQualEM - Ben Sauerwine Practice for Qualifying Exams...

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