Solution:

Ans. (a) Expression for Ampere’s circuital law $1 / 2$

Derivation of magnetic field inside the ring

$1 / 2$

(b) Identification of the material

$1 / 2$

Drawing the modification of the field pattern $1 / 2$

(a) From Ampere’s circuital law, we have,

$$ \begin{equation*} \oint \vec{B} \cdot d \vec{l}=\mu_{0} \mu_{r} I_{\text {enclosed }} \tag{i} \end{equation*} $$

For the field inside the ring, we can write

$$ \begin{aligned} & \oint \vec{B} \cdot d \vec{l}=\oint B d l=B \cdot 2 \pi r \ & \text { ( } r=\text { radius of the ring) } \ & \text { Also, } \quad I_{\text {enclosed }}=(2 \pi r n) I \end{aligned} $$

using equation (i)

$\therefore \quad B \cdot 2 \pi r=\mu_{0} \mu_{r}(2 \pi r n) I$

$\therefore \quad B=\mu_{0} \mu_{r} n I$

[Award these $\left(\frac{1}{2}+\frac{1}{2}\right)$ marks even if the result is

written without giving the derivation] $1 / 2$

(b) The material is paramagnetic. $1 / 2$

The field pattern gets modified as shown in the figure below.

$1 / 2$

[CBSE Marking Scheme 2018]

Detailed Answer :

(a)

Apply Ampere’s Law for the magnetic field due to iron ring wounded by insulating copper wire, having current I, $\oint \bar{B} \cdot \overline{d l}=\mu^{\prime} \times($ current enclosed by closed path $) \quad 1 / 2$

or, $B d l \cos 0^{\circ}=\mu^{\prime} \times(n \times 2 \times r) \times I$

or, $B \times 2 \times r \times 1=\mu^{\prime} n \times 2 \times r \times I$

or, $\quad B=\mu^{\prime} n I$

But $\quad \mu_{r}=\frac{\mu^{\prime}}{\mu_{\mathrm{o}}}$

So, $\quad B=\mu_{0} \mu_{r} n I$

This is the required expression for magnetic field. Where, $\mu_{r} \rightarrow$ relative permeability, $\mu_{0} \rightarrow$ permeability of free space, $n \rightarrow$ number of turns per unit length $n$

(b) Given : susceptibility $\chi=0.9853$ since susceptibility $\chi$ given is + ve and less than unity i.e. $\chi<+1 \quad 1 / 2$ $\Rightarrow$ magnetic material is paramagnetic material. Thus when paramagnetic material is placed in the uniform magnetic field then the modified magnetic field is shown in figure.

Paramagnetic material