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I came across this problem, and I am having a difficult time to understand the direction of E. why the final answer is in -z and y direction?, when I do the cross product I get -z and -y direction. I think I missing something here. The negative sign in front of the intrinsic impedance is throwing me off here. Can anyone help me out ? enter image description here

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To get the direction of axb (where a and b are vectors, of course ) we may use right hand rule (see image) in which forefinger represents direction of a, middlefinger represents direction of b(second vector in product) and the thumb represents the direction of product vector axb. Using this you must understand that z cross x is y but x cross z is -y. When you get this you won't find next step difficult to understand which just does - (z-y) = (-z+y).

Don't forget to use only the right hand not the left. Don't make wrong xyz coordinate axes (in which the x y z choosen by you are themselves not following x cross y equal to z).

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I think it is in the -100 (+z-y) you are having difficulty. The 100 negates the millivolts to get -1(z-y) while if you multiply the -1*z and the -1*-y you end up with (y-z) or (-z+y).

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  • \$\begingroup\$ why x cross z is -y? \$\endgroup\$
    – user65652
    Commented Sep 26, 2015 at 20:08
  • \$\begingroup\$ It's not. I just rearranged the order. -1 times +z = -z and -1 times -y = +y. hence result is (-z+y) \$\endgroup\$
    – BenG
    Commented Sep 26, 2015 at 20:13
  • \$\begingroup\$ but negative sign outside the bracket will change z and y to -z and -y, why the final answer is one negative and one positive? I don't understand this \$\endgroup\$
    – user65652
    Commented Sep 26, 2015 at 20:18
  • \$\begingroup\$ Ahh ok. I see your issue. Consider this.... -1*(z-y) is the same as -1*((+z)+(-y)) so if you expand that out you get (-1*(+z) + -1*(-y)) and further ((-z)+(+y)) which finally becomes (-z+y) \$\endgroup\$
    – BenG
    Commented Sep 26, 2015 at 20:28
  • \$\begingroup\$ oh wait. no now I see it you are referring to the second line. Sorry I'm not familar with that formula so I can't explain the -y20 or z10.... \$\endgroup\$
    – BenG
    Commented Sep 26, 2015 at 20:31

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