Shown below are pictorial and mathematical representations of an electromagnetic

Question

Shown below are pictorial and mathematical representations of an electromagnetic plane wave propagating through empty space. The electric field is parallel to the z-axis; the magnetic field is parallel to the y-axis. Recall Faraday's law, from electricity and magnetism. We shall consider how each side of the equation for Faraday's Law applies to the imaginary loop, 1 rightarrow 2 rightarrow 3 rightarrow 4 rightarrow 1, in the figure for pan A of section I. For the instant shown in the figure, determine whether each quantity below is positive, negative, or zero. Explain your reasoning in each case. the quantity evaluated over the path 1 rightarrow 2 the quantity evaluated over the path 2 rightarrow 3 the quantity evaluated over the entire loop, 1 rightarrow 2 rightarrow 3 rightarrow 4 rightarrow 1 (Hint: The answer is nor zero!) For an imaginary surface that is bounded by a closed loop, it is customary to use the right- hand rule to determine the direction of the area vector that is normal to that surface. For example, the vector that is normal to the flat, imaginary rectangular surface bounded by the loop l rightarrow 2 rightarrow 03 rightarrow 4 rightarrow 1 points in the positive y-direction.

Explanation / Answer

ITEM#1: Value of E•dl over {1 ---> 2} is positivesince path is in direction of E. Value of E•dl over {2 ---> 3} is zero since path isnormal to E. Value ofE•dl over {3 ---> 4} is negative since path is in opposite direction of E. However, diagram indicates that magnitude of E over {3 ---> 4} is less than magnitudeof E over {1 ---> 2}. Thus:      E•dl {1 ---> 2} + E•dl over {3 ---> 4} =           = {Larger Positive #} + {Smaller Negative #}           = {POSITIVE} Value of E•dl over {4 ---> 1} is zero sincepath is normal to E. In summary: E•dl {Path 1---> 2} = {POSITIVE, since path is indirection of E.} E•dl{Path 2 ---> 3} = {ZERO, since path isperpendicular to E.} E•dl {Path 1---> 2 ---> 3 ---> 4 ---> 1} =      = E•dl {1 --->2} + E•dl {2 ---> 3}+ E•dl {3 --->4} + E•dl {4 --->1}      = E•dl{1 ---> 2} + E•dl {3 --->4} + E•dl {2 --->3} + E•dl {4 --->1}      = {LargerPositive #} + {Smaller Negative #} + (0) +(0)     = {POSITIVE} Note: Above is consistent withFaraday's Law since Magnetic Flux is decreasing thru loop, which causes right-hand side of Faraday'sLaw to be positive, in agreement with closed loop {1 - 2 - 3 - 4 - 1}result. .

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