Answer to Question #207898 in Electric Circuits for Anixa

Question #207898

A coaxial cable consists of a solid inner conductor of radius a =0.2 m carrying total current of  5A, surrounded by a concentric cylindrical tube of inner radius b=0.4 m and outer radius c=0.6m carrying current of 5A. The conductors carry equal and opposite currents I distributed uniformly across their cross-sections. Determine the magnetic field at a distance r=0.5m from the axis.


1
Expert's answer
2021-06-22T10:10:05-0400

Gives

a=0.2m

Current (I)=5A

b=0.4m

c=0.6m

r=0.5m



We know that

Ampear law in closed loop

B.dl=μ0Ienclosed\oint B.dl=\mu_0I_{enclosed}

dl=2πr\oint dl=2\pi r

Ien=(II0)I_{en}=(I-I_0)

B.dl=μ0(II0)\oint B.dl=\mu_0(I-I_0)


I0=π(r2b2)π(c2b2)II_0=\frac{\pi (r^2-b^2)}{\pi(c^2-b^2)}I

I0=(r2b2)(c2b2)II_0=\frac{(r^2-b^2)}{(c^2-b^2)}I

Put value

Area π(c2b2)\pi(c^2-b^2) Flow current =II

Area 1m21m^2 Flow current=Iπ(c2b2)\frac{I}{\pi(c^2-b^2)}

Area π(r2b2)\pi(r^2-b^2) Flow current(I0)(I_0) =Iπ(c2b2)×π(r2b2)\frac{I}{\pi(c^2-b^2)}\times\pi(r^2-b^2)

B=μ0I2πr×(1r2b2c2b2)B=\frac{\mu_0I}{2\pi r}\times(1-\frac{r^2-b^2}{c^2-b^2})


Put value

B=4×3.14×107×52×3.14×0.5×(10.520.420.620.42)B=\frac{4\times3.14\times10^{-7}\times5}{2\times3.14\times 0.5}\times(1-\frac{0.5^2-0.4^2}{0.6^2-0.4^2})


B=2×106×0.55=1.1×106TB=2\times10^{-6}\times0.55=1.1\times10^{-6}T












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