a) find the instantaneous magnitude of the magnetic field at the same point and

Question

a) find the instantaneous magnitude of the magnetic field at the same point and time. The speed of light is 2.99792 X 10^8 m/s, the permeability of free space is 4pi X 10^-7 m* N/A and the permittivity of free space is 8.85419 X 10^-12 C^2/N*m^2. Answer in units of T.

b) what is the instantaneous magnitude of the poynting vector at the same point and time? Answer in units of W/m^2.

c) what is the instantaneous value of the energy density of the electric field? Answer in units of J/m^3.

d) what is the instantaneous value of the energy density of the magnetic field? Answer in units of J/m^3.

Explanation / Answer

(A) Integrate around the curve, from 0 to p radians, to sum up the total at P. Ignore the straight part, since obviously that doesn't contribute when the radial distance from the straight part is zero. I'll use the Biot-Savart law instead of Ampere's law, since there's isn't symmetry to simplify the line integral. dB = (µ0/4p) I dsxR / R^3 where x means cross product ds is the wire element vector pointing tangent to the wire. So I ds is a differential current element. r is a vector pointing toward P. Since s and r are perpendicular all along the semi-circle, the cross product is just ds * r. We rewrite dB as (µ0/4p) I ds / R² I and R are constant. So we're really just integrating ds over the wire curve from 0 to p. So all we're finding by integrating is the length of the wire that's curved. That's just pR. So B = µ0 I / 4R (B) Using the right-hand rule, it should be pretty obvious.

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