Do not forget the contribution from the straight wire at the bottom of the semic

Question

Do not forget the contribution from the straight wire at the bottom of the semicircle that runs from z = -a to z = +a. You may use the fact that the fields of the two antiparallel currents at z > a cancel.

Explanation / Answer

Because two anti parallel current at z>a will cancel each other so i wont take that part in account when calculating. -------- the curved part. There are two component of B be cause with this part. B_x=-I*pi*a*muy*sinalpha/(4pi*(x^2+a^2))=-I*a*muy*sinalpha/(4*(x^2+a^2)). where sinalpha=a/sqrt(x^2+a^2) so B_x=-I*a^2*muy*(x^2+a^2)^(-3/2)/4 about dB_y=-I*dr*cos(beta)muy*cosalpha/(4pi*(x^2+a^2)) where dr=d_beta*a so dB_y=2-I*muy*a^2*(x^2+a^2)^(-3/2)*d_beta*cos(beta)/4. integrate d_beta*cos(beta) from -pi/2 to +pi/2 we have B_y2=-I*a^2*muy*(x^2+a^2)^(-3/2)/pi2 -------- for the "diameter" part. B_y=I*muy*2*sin(theta)/4pi*z. where sin(theta)=a/sqrt(a^2+z^2). so B_y1=I*muy*a/(sqrt(a^2+z^2)*pi*2*z) ------ sum up B_y1 and B_y2 to get B_ytotal B_x = B_x total

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