%3Cp%3EA%20long%20cylindrical%20conductor%20of%20radius%20a%3D5cm%20%20carries%2

Question

%3Cp%3EA%20long%20cylindrical%20conductor%20of%20radius%20a%3D5cm%20%20carries%20a%0Acurrent%20of%2050%20amps%20into%20the%20paper%2C%20with%20a%20uniform%20current%20density.%0A%26nbsp%3BIt%20is%20surrounded%20by%20a%20shell%20(in%3Fnitely%20thin)%20of%20radius%20b%3D12cm%0Athat%20carries%20100%20amps%3C%2Fp%3E%0A%3Cp%3Eout%20of%20the%20paper%2C%20spread%20out%20uniformly%20over%20its%20surface.%20Find%0Aexpressions%20for%20the%3C%2Fp%3E%0A%3Cp%3Emagnetic%20%3Feld%20B%20(in%20Tesla)%20at%20all%20radii%2C%20giving%20both%20its%0Amagnitude%20and%20direction%2C%20in%3C%2Fp%3E%0A%3Cp%3Eterms%20of%20the%20unit%20vectors%20i%2C%20j%2C%20and%20k.%20Graph%20the%20magnitude%20of%20B%0Aas%20a%20function%20of%20r.%3C%2Fp%3E%0A%3Cp%3E(The%20z-direction%20is%20out%20of%20the%20page).%3C%2Fp%3E%0A

Explanation / Answer

here the point at which we wish to find magnetic field is inside, on the surface and outside the surface.

here we can apply cylindrical symmetry of current distribution

let Iin be the current enclosed inside the cylinder , here we must integrate current density magnitude (this is not given in question) from cyliders radius a to loop radius b.

iam assuming cuurent density relation as J = cr^2

using current density J = Iin/A

Iin =int J A

Iin = Int cr^2 (2pi rdr)

= 2pi c int r^3   fron a to b

Iin = 2pi c{r^4/4}

Iin = pi c {b^4-a^4}/2

so using amperes law

B ( 2pir) = u0{iin}

B = u0 pi c (b^4 -a^4)/2pir

B = uo pi c(b^4-a^4)/2-------------is thr formula

now soving this,

B = 4pi *10^-7 * 3,14 * (0.12^4-0l.05^4)/2

B = 3.96*10^-10 T

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