Math / Algebra

QuestionA pulse of radiation propagates with velocity vector $\vec{v}=<0,0,-c\rangle$ . The electric field in the pulse is $\vec{E}=\left\langle 4 \times 10^{6}, 0,0>N / C\right.$ . What is the magnetic field in the pulse? $\vec{B}=<0$ $\square$ $-75$ , $\square$ 0 $>T$

Studdy Solution

STEP 1

What is this asking? Given the electric field and velocity of an electromagnetic pulse, we need to find the magnetic field. Watch out! Remember the right-hand rule and the relationship between $\vec{E}$ , $\vec{B}$ , and $\vec{v}$ in an electromagnetic wave.
Don't mix up the directions!

STEP 2

1. Relationship between E, B, and v
2. Calculate B

STEP 3

Alright, so we know that in an electromagnetic wave, the electric field $\vec{E}$ , the magnetic field $\vec{B}$ , and the velocity $\vec{v}$ are all perpendicular to each other!
They form a right-handed coordinate system.

STEP 4

The relationship between their magnitudes is given by $E = cB$ , where $c$ is the speed of light.
We can rearrange this to find the magnitude of the magnetic field: $B = \frac{E}{c}$ .

STEP 5

We are given that $\vec{E} = <4 \cdot 10^6, 0, 0>$ N/C.
So, the magnitude of $\vec{E}$ is $E = 4 \cdot 10^6$ N/C.

STEP 6

The speed of light is $c = 3 \cdot 10^8$ m/s.
Now we can calculate the magnitude of B: $B = \frac{E}{c} = \frac{4 \cdot 10^6 \text{ N/C}}{3 \cdot 10^8 \text{ m/s}} = \frac{4}{3} \cdot 10^{-2} \text{ T} \approx 1.33 \cdot 10^{-2} \text{ T}$

STEP 7

Now, let's figure out the direction of $\vec{B}$ .
We know $\vec{v} = <0, 0, -c>$ and $\vec{E} = <4 \cdot 10^6, 0, 0>$ .
Using the right-hand rule (point your fingers in the direction of $\vec{E}$ , curl them towards $\vec{v}$ , and your thumb points in the direction of $\vec{B}$ ), we find that $\vec{B}$ must point in the positive y-direction.

STEP 8

Therefore, the magnetic field $\vec{B}$ is given by: $\vec{B} = <0, \frac{4}{3} \cdot 10^{-2}, 0> \text{ T} \approx <0, 1.33 \cdot 10^{-2}, 0> \text{ T}$

STEP 9

The magnetic field in the pulse is approximately $<0, 1.33 \cdot 10^{-2}, 0>$ T.

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QuestionA pulse of radiation propagates with velocity vector $\vec{v}=<0,0,-c\rangle$ . The electric field in the pulse is $\vec{E}=\left\langle 4 \times 10^{6}, 0,0>N / C\right.$ . What is the magnetic field in the pulse? $\vec{B}=<0$ $\square$ $-75$ , $\square$ 0 $>T$

Studdy Solution

STEP 1

What is this asking? Given the electric field and velocity of an electromagnetic pulse, we need to find the magnetic field. Watch out! Remember the right-hand rule and the relationship between $\vec{E}$ , $\vec{B}$ , and $\vec{v}$ in an electromagnetic wave.
Don't mix up the directions!

STEP 2

1. Relationship between E, B, and v
2. Calculate B

STEP 3

Alright, so we know that in an electromagnetic wave, the electric field $\vec{E}$ , the magnetic field $\vec{B}$ , and the velocity $\vec{v}$ are all perpendicular to each other!
They form a right-handed coordinate system.

STEP 4

The relationship between their magnitudes is given by $E = cB$ , where $c$ is the speed of light.
We can rearrange this to find the magnitude of the magnetic field: $B = \frac{E}{c}$ .

STEP 5

We are given that $\vec{E} = <4 \cdot 10^6, 0, 0>$ N/C.
So, the magnitude of $\vec{E}$ is $E = 4 \cdot 10^6$ N/C.

STEP 6

The speed of light is $c = 3 \cdot 10^8$ m/s.
Now we can calculate the magnitude of B: $B = \frac{E}{c} = \frac{4 \cdot 10^6 \text{ N/C}}{3 \cdot 10^8 \text{ m/s}} = \frac{4}{3} \cdot 10^{-2} \text{ T} \approx 1.33 \cdot 10^{-2} \text{ T}$

STEP 7

STEP 8

Therefore, the magnetic field $\vec{B}$ is given by: $\vec{B} = <0, \frac{4}{3} \cdot 10^{-2}, 0> \text{ T} \approx <0, 1.33 \cdot 10^{-2}, 0> \text{ T}$

STEP 9

The magnetic field in the pulse is approximately $<0, 1.33 \cdot 10^{-2}, 0>$ T.

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Studdy Solution

STEP 1

What is this asking? Given the electric field and velocity of an electromagnetic pulse, we need to find the magnetic field. Watch out! Remember the right-hand rule and the relationship between E⃗\vec{E}E, B⃗\vec{B}B, and v⃗\vec{v}v in an electromagnetic wave.Don't mix up the directions!

STEP 2

1. Relationship between E, B, and v2. Calculate B

STEP 3

Alright, so we know that in an electromagnetic wave, the electric field E⃗\vec{E}E, the magnetic field B⃗\vec{B}B, and the velocity v⃗\vec{v}v are all perpendicular to each other!They form a right-handed coordinate system.

STEP 4

The relationship between their magnitudes is given by E=cBE = cBE=cB, where ccc is the speed of light.We can rearrange this to find the magnitude of the magnetic field: B=EcB = \frac{E}{c}B=cE​.

STEP 5

We are given that E⃗=<4⋅106,0,0>\vec{E} = <4 \cdot 10^6, 0, 0>E=<4⋅106,0,0> N/C.So, the magnitude of E⃗\vec{E}E is E=4⋅106E = 4 \cdot 10^6E=4⋅106 N/C.

STEP 6

STEP 7

STEP 8

Therefore, the magnetic field B⃗\vec{B}B is given by: B⃗=<0,43⋅10−2,0> T≈<0,1.33⋅10−2,0> T \vec{B} = <0, \frac{4}{3} \cdot 10^{-2}, 0> \text{ T} \approx <0, 1.33 \cdot 10^{-2}, 0> \text{ T} B=<0,34​⋅10−2,0> T≈<0,1.33⋅10−2,0> T

STEP 9

The magnetic field in the pulse is approximately <0,1.33⋅10−2,0><0, 1.33 \cdot 10^{-2}, 0><0,1.33⋅10−2,0> T.

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What is this asking? Given the electric field and velocity of an electromagnetic pulse, we need to find the magnetic field. Watch out! Remember the right-hand rule and the relationship between $\vec{E}$ , $\vec{B}$ , and $\vec{v}$ in an electromagnetic wave.
Don't mix up the directions!

1. Relationship between E, B, and v
2. Calculate B

Alright, so we know that in an electromagnetic wave, the electric field $\vec{E}$ , the magnetic field $\vec{B}$ , and the velocity $\vec{v}$ are all perpendicular to each other!
They form a right-handed coordinate system.

The relationship between their magnitudes is given by $E = cB$ , where $c$ is the speed of light.
We can rearrange this to find the magnitude of the magnetic field: $B = \frac{E}{c}$ .

We are given that $\vec{E} = <4 \cdot 10^6, 0, 0>$ N/C.
So, the magnitude of $\vec{E}$ is $E = 4 \cdot 10^6$ N/C.

Therefore, the magnetic field $\vec{B}$ is given by: $\vec{B} = <0, \frac{4}{3} \cdot 10^{-2}, 0> \text{ T} \approx <0, 1.33 \cdot 10^{-2}, 0> \text{ T}$

The magnetic field in the pulse is approximately $<0, 1.33 \cdot 10^{-2}, 0>$ T.