Question

In: Physics

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the...

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave?

A)\vec Ein the +x direction,\vec Bin the +y direction.Please Choose+ x direction- x direction+ y direction- y direction+ z direction- z direction

B)\vec Ein the -y direction,\vec Bin the +x directionPlease Choose+ x direction- x direction+ y direction- y direction+ z direction- z direction

C)\vec Ein the +z direction,\vec Bin the -x direction.Please Choose+ x direction- x direction+ y direction- y direction+ z direction- z direction

D)\vec Ein the +y direction,\vec Bin the -z direction.Please Choose+ x direction- x direction+ y direction- y direction+ z direction- z direction

Solutions

Expert Solution

Concepts and reason

The concepts required to solve the given questions is the propagation of wave.

Initially, determine the direction of the electric field and the magnetic field vector from the given question. Later, write a formula for the direction calculation of the wave propagation. Finally, find the direction of propagation of the wave.

Fundamentals

The expression for the direction of propagation of wave is,

d=E×B\vec d = \vec E \times \vec B

Here, d\vec d is the direction, E\vec E is the electric field vector, and B\vec B is the magnetic field vector.

(A)

Substitute E(+x^)E\left( { + \hat x} \right) for E\vec E and B(+y^)B\left( { + \hat y} \right) for B\vec B in the equation d=E×B\vec d = \vec E \times \vec B .

d=E(+x^)×B(+y^)=EB(+x^)×(+y^)=EB(+z^)\begin{array}{c}\\\vec d = E\left( { + \hat x} \right) \times B\left( { + \hat y} \right)\\\\ = EB\left( { + \hat x) \times ( + \hat y} \right)\\\\ = EB\left( { + \hat z} \right)\\\end{array}

(B)

Substitute E(y^)E\left( { - \hat y} \right) for E\vec E and B(+x^)B\left( { + \hat x} \right) for B\vec B in the equation d=E×B\vec d = \vec E \times \vec B .

d=E(y^)×B(+x^)=EB(y^)×(+x^)=EB(+z^)\begin{array}{c}\\\vec d = E\left( { - \hat y} \right) \times B\left( { + \hat x} \right)\\\\ = EB\left( { - \hat y) \times ( + \hat x} \right)\\\\ = EB\left( { + \hat z} \right)\\\end{array}

(C)

Substitute E(+z^)E\left( { + \hat z} \right) for E\vec E and B(x^)B\left( { - \hat x} \right) for B\vec B in the equation d=E×B\vec d = \vec E \times \vec B .

d=E(+z^)×B(x^)=EB(+z^)×(x^)=EB(y^)\begin{array}{c}\\\vec d = E\left( { + \hat z} \right) \times B\left( { - \hat x} \right)\\\\ = EB\left( { + \hat z) \times ( - \hat x} \right)\\\\ = EB\left( { - \hat y} \right)\\\end{array}

(D)

Substitute E(+y^)E\left( { + \hat y} \right) for E\vec E and B(z^)B\left( { - \hat z} \right) for B\vec B in the equation d=E×B\vec d = \vec E \times \vec B .

d=E(+y^)×B(z^)=EB(+y^)×(z^)=EB(x^)\begin{array}{c}\\\vec d = E\left( { + \hat y} \right) \times B\left( { - \hat z} \right)\\\\ = EB\left( { + \hat y) \times ( - \hat z} \right)\\\\ = EB\left( { - \hat x} \right)\\\end{array}

Ans: Part A

The direction of propagation of wave is in the positive z direction, that is +z^+ \hat z .

Part B

The direction of propagation of wave is in the positive z direction, that is +z^+ \hat z .

Part C

The direction of propagation of wave is in the positive y direction, that is y^- \hat y .

Part D

The direction of propagation of wave is in the negative x direction, that is x^- \hat x .


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