QUIZ9_SOLN

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ECE 325 Miami University Quiz 9 (Take-Home) Name:__SOLUTIONS_____ Due by email: November 19, 2021 Tasks: 1. A loop made of flexible wire with current I flowing through it is placed in magnetic field as shown below (the field is pointing into the page). Using the knowledge of magnetic force, describe what happens to the loop in this case and also what would happen to it if the current direction was reversed. ( 10 points ) I B Ans: Using the right-hand rule (current going into the palm, field is aligned with the fingers, then force points in the direction of the thumb), we can see that for every segment of the loop the magnetic force F will be pointing inward – thus, the loop will bend inwards. When current direction is reversed, the loop will expand outward. 2. Shown below is a basic scheme of a “can crusher” (but without a magnet). At time 0 the key is closed and a high-voltage capacitor sends current I into the wire around the can, as shown. In order to create a “crushing” force, however, you also need a source of external magnetic field – e.g. a bar magnet. Sketch a bar magnet with both poles (N and S) clearly shown and positioned in such a way as to t urn this into a true “can crusher.” ( 10 points ).

ECE 325 Miami University I C + - Ans: For the magnetic force to point inward at every loop, we need the magnetic field to point “ down the can, ” as shown in the picture above. This could be accomplished, e.g., by placing a magnet as shown. 3. Review the problem about boundary conditions for the electric field. Assuming the same geometry (two media separated by a boundary – see below), and using Ampère ’s Circuital Law in integral form, derive boundary conditions for the magnetic field H using similar setup for the field lines. Hint: Assume that the only current that can flow in Medium 2 (bottom medium) is the one parallel to the boundary and in the direction from point a to point b and no current can flow in Medium 1. ( 15 points ). Ans: Follow the solution for a similar boundary problem in Electrostatics. Replacing the E-field equation with Ampère ’s Circuital Law , we obtain: enc abcd H dl I • =  , where I enc is the current going through the abcd loop. The problem states that the only current flowing in this scenario is Medium 1 Medium 2 H 1 H 2

ECE 325 Miami University going from a to b , i.e., it does not flow through the loop, thus the right-hand side of the above equation turns to zero. To obtain the solution, follow exact same steps we did in class when solving the same problem in Electrostatics: Break the H-field down into normal and tangential components, then write out four integrals for the four sides of the rectangular loop abcd and solve them, taking care to find the dot-products correctly. Once all four integrals are summed up, the result will be identical to the one we obtained in Electrostatics, only written in terms of the magnetic field H: 1 2 T T H H = , i.e., the tangential components of the field just above and below the boundary are equal. 4. Using a simple illustration, briefly explain the “rule of thumb” regarding an antenna length that has to be at least equal to half the signal’s wavelength for efficient transmission or reception of that signal. ( 15 points ). Ans: The positive half wave of the input signal from the generator will cause the current to flow “ up ” the antenna (think of it in terms of the field, though – it ’ s the E-field that will propagate upwards in the antenna) – until it reaches the end, at which time it will cease flowing (recall the “ Cable to the Moon ” problem). Assuming the field/current propagates with speed of light c , it ’ s easy to figure out that the length of the antenna L must be not shorter than 2 cT for the current/field to exist in the antenna over the entire duration of the input signal from a generator. Recalling that c f  = and 1/ f T = , we conclude that 2 L   . 5. Follow steps below to arrive at the solution for magnetic field inside a capacitor in AC conditions. ( 50 points )

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ECE 325 Miami University a. Assume a parallel plate capacitor with separation between the plates equal h (as illustrated above). It can be shown using Electrostatics, that the electric potential inside the capacitor is equal to 0 ( ) V V z z h = , where V 0 is voltage amplitude on the capacitor. Assume that voltage amplitude is a function of time: ( ) 0 0 ( ) cos 2 V t A f t  = and, using the gradient expression, find the electric field inside this capacitor: E V = − . (We can still use the expressions above in quasi-static approximation). Hint: Note that V(z) is function of z only, so you can use the expression for gradient in the Cartesian coordinate system for this step: x y z a a a x y z     = + +    . Ans: V only depends on z, so the only non-zero term of the gradient will be the last one, thus ( ) 0 0 0 cos 2 z z z V z A f t V h E a a a z h h         = − = − = −  b. Now, consider the Ampère-Maxwell equation in differential form: 0 r E H E t      = +  . Assume that the capacitor has air inside, so that  = 0 and  r = 1. Using the result from Step a above, find the right-hand side of this equation. Ans: Applying the time derivative to the result obtained in Step 1 above, we obtain the right-hand side of the equation as: ( ) ( ) 0 0 0 0 0 0 cos 2 2 sin 2 r r r z z A f t h f A E a f t a t t h                  = − =  

ECE 325 Miami University c. Briefly explain why the magnetic field inside the capacitor has only the  -component (see drawing below for the coordinate system explanation). Ans: We can use Biot-Savart ’ s Law, where the current I is replaced by the displacement current E t   (as discussed in lecture). Since this current is directed along the z-coordinate only, the H-field at every point inside the capacitor will be pointing along the   -coordinate due to the right-hand rule. d. Using the fact that ( ) H H t a  = , find the left-hand side expression by taking curl of H in cylindrical coordinates. Hint: The curl in cylindrical coordinate system is ( ) 1 1 z z z A A A A A A A a a a z z                             = − + − + −                   . Note that the vector components on the left and on the right sides of the resultant Ampère- Maxwell equation must match. Ans: Notice that the curl equation contains all three vector components of a general vector field A ( , , z   ) . In our case, however, the magnetic field of interest has only the  vector

ECE 325 Miami University component – thus, all terms containing A  and A z components will automatically turn to zeroes, leaving only two potentially non-zero terms: ( ) 1 1 z z z A A A A A A H a a a z z                             = − + − + −                   where A  will be replaced with H-field  -component ( ) H t . H is assumed to have no z-dependence, so the first surviving term will go to zero as well. The second term will survive because it gets multiplied by  , so the result will be: ( ) z H t H a   = e. By equating the left and the right sides of the resultant Ampère-Maxwell equation, find H(t) inside the capacitor and briefly discuss. Ans: Equating the right and left sides gives us: ( ) 0 0 0 2 ( ) sin 2 r z z f A H t a f t a h      = , which allows us to solve for H(t) immediately – ( ) 0 0 0 2 ( ) sin 2 r f A H t f t h      = . We notice the following: • H-field in 90-degrees out of phase with the E-field • H-field ’ s magnitude is directly proportional to the signal ’ s frequency • At DC (statics) H-field is zero • H-field is inversely proportional to the distance between capacitor ’ s plates • H-field is directly proportional to the distance  … … which leads us to the last bonus step. f. Re-solve the curl equation for  -dependent H. Ans: Note that Step d started with an assumption that ( ) H H t a  = - i.e., we assumed that H is only dependent upon time and nothing else. This assumption is disproved by our own solution, which shows that H depends on  . Let ’ s plug this back into the curl equation in the new form ( ) H H t a   = and re-do the operation: ( ) ( ) 2 ( ) 1 ( ) 2 ( ) 2 ( ) z z z z H t H t H t H a a a H t a                   = = = =           Equating the right- and left-hand sides of the Ampère-Maxwell equation once again, we solve for H(t) :

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ECE 325 Miami University ( ) 0 0 0 ( ) sin 2 r f A H t f t h     = , which then leads us to our final and correct expression for the H-field (remember, its new form is ( ) H H t a   = , so H(t) must now be multiplied by  ): ( ) 0 0 0 sin 2 r f A H f t a h       =