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a pulse of radiation propagates with velocity vector v→v→ = <0, 0, −c-c>. the electric field in the pulse is e→=e→= <8.3 × 106, 0, 0> n/c. what is the magnetic field in the pulse?

User Jaquann
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Final answer:

The magnetic field in the electromagnetic pulse is perpendicular to both the electric field and direction of propagation. Calculating the magnitude of the magnetic field gives us B = 0.0277 T, and following the right-hand rule, the magnetic field vector is B = <0, –0.0277, 0> T.

Step-by-step explanation:

According to Maxwell's equations and the properties of electromagnetic waves, the electric field E and the magnetic field B are always perpendicular to each other in a propagating electromagnetic wave, and both are perpendicular to the direction of propagation of the wave. This relationship is captured by the equation B = E / c, where c is the speed of light in a vacuum, which is approximately 3 × 10^8 m/s.

In the specific case given, the electric field is E = <8.3 × 10^6, 0, 0> N/C, and the velocity vector of propagation is v = <0, 0, –c>. Since B must be perpendicular to both E and v, and has a magnitude given by |E|/c, which is 8.3 × 10^6 N/C divided by 3 × 10^8 m/s, we can calculate the magnitude of the magnetic field to be B = 0.0277 T (Tesla). The direction of B must be such that E, B, and v follow the right-hand rule. Since E is in the x-direction and v is in the negative z-direction, B must be in the negative y-direction to satisfy this rule. Therefore, the magnetic field in the pulse is B = <0, –B, 0> T, or more specifically B = <0, –0.0277, 0> T.

User Ollie Strevel
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