Question

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

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

r = distance of a point from centre
For r $\le$ R/2 Using Ampere's circuital law,
$\oint B.d\mathbf{1} \, \, \, \, \, or \, \, \, \, \, \, \, Bl=\mu_0(I_{in})$
$or \, \, \, \, \, \, \, B(2\pi r)\, \, \, \, \, or\, \, \, \, \, \, \, \, Bl=\mu_0(I_{in})\, or\, B=\frac{\mu_0}{2\pi}\frac{I_{in}}{r}\, \, \, \, \, \, \, \, ...(i)$
Since, $I_{in}=0\Rightarrow \, \, \, \, \, \therefore B=0$
$For \frac{R}{2} \le r \le R\, \, \, \, \, \, \, I_{in}=\Bigg[\pi r^2 -\pi \Bigg(\frac{R}{2}\Bigg)^2\Bigg]\sigma$
Here$\sigma$ = current per unit area.
Substituting in Eq. (i), we have
$B=\frac{\mu_0}{2\pi}\frac{\Bigg[\pi r^2 -\pi \frac{R^2}{4}\Bigg]\sigma}{r}=\frac{\mu_0 \sigma}{2r}\frac{\mu_0}{2\pi}\Bigg(r^2-\frac{R^2}{4}\Bigg)$
At$\, \, \, \, \, \, r=\frac{R}{2}, B=0$
At$\, \, \, \, \, \, r=R, B=\frac{3\mu_0 \sigma R}{8}$
For e $\ge R\, \, \, \, \, \, \, \, \, \, I_{in}=I_{Total}=I$(say)
Therefore, substituting in Eq. (i), we have
$B=\frac{\mu_0}{2\pi}.\frac{I}{r}\, \, \, \, \, \, or \, \, \, \, \, \, B \propto \frac{1}{r}$