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Tutorial 1 11th August 2008 + Answer 1. Consider the rotating conductor shown. The center of the 2a diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make conductive contact with a ring to make one end of the electrical contact to R; the other contact is made to the center of the bar. Given Bo = 100 mWb/m2, a = 6 cm, and R = 50 , determine I if the bar rotates at 2π radians per second. Figure indicates one of the paths for the circulation integral. a a Vemf a Boa z d a Bo d 0 I Vemf R 0 Bo a Bo a 2 2 1 rev rad 2 1 Vs A 3 Wb 1 0.06m 2 100 x10 2 2R 2 s rev m 50 Wb V 2 I 22.6 A 2. A conductive rod, of length 6.0 cm, has one end fixed on a grounded origin and is free to rotate in the x-y plane. It rotates at ω = 120π radian per second in a magnetic field B = 100 mT az. Find the voltage at the end of the bar. 60 u rev 2 rad rad 120 s rev s d a a dt l Vemf a Boa z d a Bo l 2 2 0 1 rad Wb 2 Vs Vemf 120 0.1 2 0.06m 2 s m Wb 68mV We can confirm the sign by observing that a positive charge placed in the middle of the bar would move to the ungrounded end by the Lorentz force equation. 3. A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer conductor diameter 6.0 mm is filled with an ideal dielectric with r = 10.2. A voltage v(t) = 10.cos(6x106 t) mV is placed on the inner conductor and the outer conductor is grounded. Neglecting fringing fields at the ends of the coax, find the displacement current between the inner and outer conductors. Capacitance for coaxial cable is given by: 2 lVo cos t 2 l for coax (from chapter 2): C= so Q ln b a ln b a C Q , so Q Cv(t ) V 2 l Vo cos t Vo cos t V cos t and D o a . 2 l ln b a ln b a ln b a D Vo sin t a Jd t ln b a so D 2 l Vo sin t 1 2 l Vo sin t id J d dS d dz ln b a 0 ln b a 0 Now evaluate id with the given parameters: 2 6 x106 rad s 10.2 8.854 x10 12 F m 1m 0.010V sin t C As id ln 0.006 0.002 FV C id 97 sin 6 x106 t A 4. A magnetic field propagating in free space is given by H( z , t ) 20.sin 108 t z a x A m . Find frequency, phase constant, wavelength and the electric field, E(z,t). 1 x108 50MHz 2 8 c 3x10 2 rad 6m, 8 f 0.5x10 3 m D H (no J c ) t 2 f x108 , f ax H x H o sin t z H o cos t z a y o dE E = Ho ay y 0 az z 0 E t cos t z dta o Ho V sin t z a y so E( z, t ) o m y z Ho o H o sin t z a y sin t z a y Inserting the given values we find: E( z, t ) 7.5sin x108 t 3 kV z ay m 5. A 10.0 MHz magnetic field travels in a fluid for which the propagation velocity is 1.0x108 m/sec. Initially, we have H(0,0)=2.0 ax A/m. The amplitude drops to 1.0 A/m after the wave travels 5.0 meters in the y direction. Find the general expression for this wave. The general expression for the wave is: A H( y , t ) H o e y cos t y a x m rad 2 f 2 10 x106 20 x106 s u 2 rad u p 1x108 m s ; p 10m; 0.2 f m A H(0,0) 2.0a x ; H o 2.0, 0 m H ( y 5) 1 H oe y 2.0e (5) ; solving we get 0.14 Finally, H ( y, t ) 2.0e 0.14 y cos 20 x106 t 0.2 y a x A m 6. In free space, E( z, t ) 10.cos 106 t z a x 20.cos 106 t z a y V . m Find H(z,t) using phasors. The phasor version of E is: Es 10e j z a x 20e j z a y ax Es x 10e j z ay y 20e j z az z 0 20e j z a x 10e j z a y z z Es j 20e j z a x j 10e j z a y jo Hs j 20 j z j 10 j z 20 j z 10 j z Hs e ax e ay e ax e ay jo jo o o up x106 rad ; 10.47 x103 8 u p 3x10 m Inserting the appropriate values we then find: mA H s 53e j z a x 27e j z a y , m and H( z, t ) 53cos t z a x 27 cos t z a y mA m