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alexira [117]
3 years ago
13

An electromagnetic wave is propagating in the positive x direction. At a given moment in time, the magnetic field at the origin

points in the positive y direction. In what direction does the electric field at the origin point at that same moment?a. Positive xb. Negative xc. Positive yd. Negative ye. Positive zf. Negative z
Physics
1 answer:
Flauer [41]3 years ago
8 0

To solve this exercise it is necessary to take into account the concepts related to the magnetic field and its vector representation through the cross product or vector product.

According to the definition the direction of the electromagnetic wave propagation is given by

\hat{n} = \vec{E} \times \vec{B}

Where,

E = Electric Field

B = Magnetic Field

According to the information provided, the direction of propagation of the electromagnetic wave is on the X axis, which for practical purposes we will denote as \hat {i}, on the other hand it is also indicated that the magnetic field is in the Y direction, that for practical purposes we will denote it as \hat {j}. In this way using the previous equation we would have to,

\hat{n} = \vec{E} \times \vec{B}

\hat{i} = \vec{E} \times \hat{j}

The cross product identity is

\hat{i}=-\hat{k} \times \hat{j}

From the equation we can notice that the electric field would be given by,

\vec{E} = -\hat{k}

<em>Therefore the direction of electric field is negative z-axis. </em>

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Vladimir79 [104]
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Answer: 27000 km/s
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4 years ago
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Explanation:

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5 0
4 years ago
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Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

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</span><span>
</span>
5 0
3 years ago
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