Please write a clear and concise solution as i am desperate...

Please write a clear and concise solution as I am desperate to understand. Please show all steps. Please solve all parts a, b, and c, including the sketch. Use H instead of B for the magnetic field. Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude E0, frequency ω, and phase angle φ that is (a) Traveling in the -x direction with electric field polarized in the z direction (b) Traveling in the direction from the origin to the point (1,1,1) with electric field polarized parallel to the xz plane. (c) Sketch both waves and give the explicit Cartesian components of the unit vectors hat(k) and hat(E). For the wave from part a, it will be easiest if the three axes are x, y, and z. For the wave from part b, it will be easiest if one of the axes is in the direction of hat(k) (the [111] direction), another is in the direction of hat(E), and the third is in the direction of hat(H).

Transcript

00:01 Hello students here is the solution.

00:03 Here given that an electromagnetic wave which have two field, electric field and magnetic field are given in the question and the electric field is this and the magnetic field is this.

00:12 In a rectangular loop, 1, 2, 3, 4, this is the loop p is lie in x -z so first of all here we have to find out the close integration of e .d .m.

00:21 So, students, first of all, we know that, we know that we know that k is equals to 2 pi divided by lambda from this equation lambda is equal to 2 pi divided by k and from here the lambda divided by 2 or the half of the wavelength is equal to pi divided by k this is the first relation of this question and here we have to find out the close integration of e dot d l so e tot d l is equal to integration e dot d l we will convert or we will divide it into four parts one two three and four so first of all this the integration of part 1 plus integration of part 2 which is e dot dl and this is the second line plus integration e .d .l plus integration 4 e .d .l vector.

01:06 So dear students, here, we know that here, we know that there is no variation.

01:14 There is no variation along z -axis.

01:23 There's no variation along the x -axis.

01:26 So that's why here we can say that that this term and this term will be zero.

01:30 So here the integration e .d .l or close integration e .d .l is equal to integration of e .dl from first plus integration e dot d l from second now put the values here and after that we can write it as that will be equals to integration 0 to h e cos k z minus omega t plus integration h2 0 e cos k z minus omega t d x and here also the d x in this term from 0 to h the z will be 0 and from h2 0 the value of z is lambda by 2 so dearest here z equals to 0 and here z is equals to 5 divided by k k and here lambda divided by 2 is equal to pi divided by k now put all the values here and after that we get close integration e dot d l is equal to e integration 0 to h cost so i put the value 0 in this equation and we get cos omega t bx plus integration h2 0 e outside from this equation because e is a constant and after put the values that is equals to lambda divided by k we get that cost by minus omega t multiply by d x and after solving this equation after the integration we get that eh cause omega t plus eh cos omega t so now the value of cross integration e .d .l is equal to twice of eh cos omega t and that is the answer of the first part and that is the answer of the first part now moving to the second part in second part we have to find out the magnetic flux through this loop so in this second part we have to find out the magnetic flux through loop b okay so we know that we know that we know that the magnetic flux 5 is equals to integration of b .da, right? and here we know that that da is equals to minus y gap dx.

03:59 That's because here this little part is dz and this little part is dx and the area vector along the minus y at this.

04:09 So here, d .a is equals to minus y gap bx dot dz.

04:13 Now put the value here, so it is the double integration and here the minus of double integration b dot y cap bx b z now put all the values here that will be equals to minus integration 0 to h because the limit of x is from 0 to h and the limit of z is from 0 to pi divided by k and here cos k z minus omega t b x bx multiply by d z first of all integrate with respect to x so here they integrate with respect to x and we get the value here that minus of b h integration 0 to pi divided by k k cos k z minus omega t d z i'm integrated with respect to d z and we get the value here that minus v h as it is and the integration of course k z minus omega t here z is the variable so here the integration is sign k z minus omega t limit from 0 to pi divided by k now put the values here so it will be minus of bh divided by k so integration is also divided by k so minus of bh divided by k sine pi minus omega t minus sine minus of omega t after solving this equation we get the value of phi or the magnetic flux is equal to minus of 2 bh divided by k sine omega t so this is the answer of the second part now moving to the third part in part c we have to to find out the relation between b, e, and c from the faradish l.

05:52 So dear students, here is the solution of the part c.

05:55 So here from faraday's law, so here from faraday's law, so from faraday's law, we know that, that the clause integration, d.

06:05 D .l, is equal to minus of d5 divided by dt.

06:09 So here, d5 divided by dt is equals to minus 2bh omega divided by k, cos omega t.

06:18 But we know that, that omega, that omega, divided by k is equals to c so here we can say that that minus of b5 by d t is equals to so here minus d phi by d t is equals to my 2 b h c c c c then we can write it as that 2 e h cost omega t is equal to twice of b h c c c c c from this two equation here we can write it as that e is equals to b multiplier by c or b is equal to e divided by c.

06:53 So this is the answer of the third part...

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