Exam3_Spring2007_Conflict_sol_02

Exam3_Spring2007_Con - 8.02 Conflict Exam Three Spring 2007 Solutions PART I(25 points Question A(5 points A parallel-plate capacitor with circular

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1 8.02 Conflict Exam Three Spring 2007 Solutions PART I (25 points) Question A (5 points). A parallel-plate capacitor with circular plates of radius R and separated by a distance d is discharged through a straight wire carrying current I , as shown in the figure below: During the time interval that the plates are discharging, the direction of the magnetic field at the point P on the figure above 1. points into the page 2. points out of the page 3. points radially outward away from the axis of the capacitor. 4. points radially inward toward the axis of the capacitor. 5. points from bottom to top 6. points from top to bottom Solution: 1 . A time changing electric field acts like a current with regards to creating magnetic field. The direction of the magnetic field depends on two factors, (1) which direction the electric field is pointing and (2) whether it is increasing or decreasing. The generalized Ampere’s Law is 00 0 through Amperian open loop surface d dI d dt µµ ε ⋅= + ∫∫ Bs Ea GG G G v . (1.1)
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2 If we choose a circle centered on the axis in between the plates as our Amperian loop, and the disk defined by the circle as the open surface then there is no conduction current flowing through the disk, only changing electric flux. Then Eq (1.1) becomes 00 circle disk d dd dt µε =⋅ ∫∫ Bs Ea G G G G v . (1.2) (This is case is similar to determining the direction of the electric field when magnetic flux is changing, however Faraday’s Law has a minus sign which will give directions opposite to our case). We orient the disk, so that the electric flux is positive, this means the circle is oriented counterclockwise as seen from above. Since the plates are discharging, the electric field is decreasing, hence the electric flux is decreasing. Therefore, the line integral is negative so that the direction of the magnetic field must point opposite the direction that we circulated around the line integral which was counterclockwise (as seen from above). Thus at the point P on the left, the magnetic field points into the plane of the paper.
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3 Question B (5 points). Consider a plane EM wave with magnetic field given by 0 ˆ cos( ) B kx t ω =+ Bk G (1.3) Which is the corresponding electric field? 1. 0 ˆ cos( ) cB kx t Ek G . 2. 0 ˆ cos( ) cB kx t =− + G . 3. 0 ˆ cos( ) B kx t c G . 4. 0 ˆ cos( ) B kx t c + G . 5. 0 ˆ cos( ) cB kx t E j G . 6. 0 ˆ cos( ) cB kx t + E j G . 7. 0 ˆ cos( ) B kx t c E j G . 8. 0 ˆ cos( ) B kx t c + E j G . 9. 0 ˆ cos( ) cB kx t Ei G . 10. 0 ˆ cos( ) cB kx t + G . 11. 0 ˆ cos( ) B kx t c G . 12. 0 ˆ cos( ) B kx t c + G . Solution 6. The amplitude of the electric field is larger by a factor of the speed of light from the amplitude of the electric field, hence 00 Ec B = . Recall that at 0 x = and 0 t = , cos(0) 1 = , so the magnetic field is maximum on the plane 0 x = . For 0 t > , the argument of the cosine function remains zero when 0 kx t + = . Therefore the maximum at 0 x = and 0 t = propagates according to ( / ) x kt c t = −= which is in the negative
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4 x - direction. The electric field must have the same sinusoidal form, indicating that it is also traveling in the negative x - direction. The direction of the electric and magnetic
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This note was uploaded on 04/27/2009 for the course 8 8.02 taught by Professor Hudson during the Spring '07 term at MIT.

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Exam3_Spring2007_Con - 8.02 Conflict Exam Three Spring 2007 Solutions PART I(25 points Question A(5 points A parallel-plate capacitor with circular

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