- #1
lam58
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For question 4b in the problem sheet attached below, is my working correct for B outside and inside the coaxial cable?
Ans:
Outside coaxial:∫B.dl = μI (the integral is between 0 and 2∏r)
=> B = (μI)/(2∏r)
Inside the coaxial: [tex]\int_{2\pi a}^{2\pi r} B.dI[/tex]
= [tex] μI.\frac{a^2}{\pi r^{2}b^{2}}[/tex]
=> [tex]B[2\pi r - 2\pi a] = μI.\frac{a^2}{\pi r^{2}b^{2}}[/tex]
=>[tex]B = μI.\frac{a^2}{\pi r^{2}b^{2}} * \frac{1}{2\pi r - 2\pi a}[/tex]
Ans:
Outside coaxial:∫B.dl = μI (the integral is between 0 and 2∏r)
=> B = (μI)/(2∏r)
Inside the coaxial: [tex]\int_{2\pi a}^{2\pi r} B.dI[/tex]
= [tex] μI.\frac{a^2}{\pi r^{2}b^{2}}[/tex]
=> [tex]B[2\pi r - 2\pi a] = μI.\frac{a^2}{\pi r^{2}b^{2}}[/tex]
=>[tex]B = μI.\frac{a^2}{\pi r^{2}b^{2}} * \frac{1}{2\pi r - 2\pi a}[/tex]