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Physics Question on Moving charges and magnetism

An infinitely long hollow conducting cylinder with inner radius R/2R/2 and outer radius RR carries a uniform current density along its length. The magnitude of the magnetic field, as a function of the radial distance rr from the axis is best represented by

A

B

C

D

Answer

Explanation

Solution

r = distance of a point from centre
For r \le R/2 Using Ampere's circuital law,
B.d1orBl=μ0(Iin)\oint B.d\mathbf{1} \, \, \, \, \, or \, \, \, \, \, \, \, Bl=\mu_0(I_{in})
orB(2πr)orBl=μ0(Iin)orB=μ02πIinr...(i)or \, \, \, \, \, \, \, B(2\pi r)\, \, \, \, \, or\, \, \, \, \, \, \, \, Bl=\mu_0(I_{in})\, or\, B=\frac{\mu_0}{2\pi}\frac{I_{in}}{r}\, \, \, \, \, \, \, \, ...(i)
Since, Iin=0B=0I_{in}=0\Rightarrow \, \, \, \, \, \therefore B=0
ForR2rRIin=[πr2π(R2)2]sFor \frac{R}{2} \le r \le R\, \, \, \, \, \, \, I_{in}=\Bigg[\pi r^2 -\pi \Bigg(\frac{R}{2}\Bigg)^2\Bigg]s
Heress = current per unit area.
Substituting in E (i), we have
B=μ02π[πr2πR24]sr=μ0s2rμ02π(r2R24)B=\frac{\mu_0}{2\pi}\frac{\Bigg[\pi r^2 -\pi \frac{R^2}{4}\Bigg]s}{r}=\frac{\mu_0 s}{2r}\frac{\mu_0}{2\pi}\Bigg(r^2-\frac{R^2}{4}\Bigg)
Atr=R2,B=0\, \, \, \, \, \, r=\frac{R}{2}, B=0
Atr=R,B=3μ0sR8\, \, \, \, \, \, r=R, B=\frac{3\mu_0 s R}{8}
For e RIin=ITotal=I\ge R\, \, \, \, \, \, \, \, \, \, I_{in}=I_{Total}=I(say)
Therefore, substituting in E (i), we have
B=μ02π.IrorB?1rB=\frac{\mu_0}{2\pi}.\frac{I}{r}\, \, \, \, \, \, or \, \, \, \, \, \, B?\frac{1}{r}