Magnetics Ques 67
- 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, |B| as a function of the radial distance $r$ from the axis is best represented by
(2012)
(a)
(b)
(c)
(d)
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Answer:
Correct Answer: 67.(d)
Solution:
$r=$ distance of a point from centre.
For $\boldsymbol{r} \leq \boldsymbol{R} / \mathbf{2}$ Using Ampere’s circuital law,
$ \begin{aligned} & g \mathbf{B} \cdot d \mathbf{l} \text { or } B l=\mu _0\left(I _{in}\right) \\ & B(2 \pi r)=\mu _0\left(I _{\text {in }}\right) \text { or } \quad B=\frac{\mu _0}{2 \pi} \frac{I _{\text {in }}}{r} \quad …….(i) \end{aligned} $
Since, $I _{\text {in }}=0 \Rightarrow \therefore B=0$
For $\frac{\boldsymbol{R}}{\mathbf{2}} \leq \boldsymbol{r} \leq \boldsymbol{R} \quad I _{\text {in }}=[\pi r^{2}-\pi (\frac{R}{2})^{2}] \sigma$
Here, $\sigma=$ current per unit area
Substituting in Eq. (i), we have
$ B=\frac{\mu _0}{2 \pi} \frac{[\pi r^{2}-\pi \frac{R^{2}}{4} ]\sigma}{r}=\frac{\mu _0 \sigma}{2 r} (r^{2}-\frac{R^{2}}{4}) $
At $\quad r=\frac{R}{2}, B=0$
At $\quad r=R, B=\frac{3 \mu _0 \sigma R}{8}$
For $\boldsymbol{r} \geq \boldsymbol{R} \quad I _{\text {in }}=I _{\text {Total }}=I$ (say)
Therefore, substituting in Eq. (i), we have
$ B=\frac{\mu _0}{2 \pi} \cdot \frac{I}{r} \quad \text { or } \quad B \propto \frac{1}{r} $
BETA
