Solution:

Ans. Formula for magnetic field of toroid 1

Calculation of magnetic field $1 \frac{1}{2}$

Effect of change of core

$1 / 2$ $1 / 2$

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

$$ =\left(800 \times 4 \pi \times 10^{-7}\right) \times\left(\frac{4000}{2 \pi \times 20 \times 10^{-2}}\right) \times 3 $$

$$ =9.6 \mathrm{~T} $$

Since Bismuth is diamagnetic, its $\mu_{r}<1$

$\therefore$ The magnetic field in the core will get very much reduced.

[CBSE Marking Scheme 2017]

Detailed Answer :

Given :

Mean radius of toroidal solenoid $=20 \mathrm{~cm}$

Number of turns of wire wound $=4000$

Relative permeability of ferromagnetic core $=800$

Current passing through the coil $=3 \mathrm{~A}$

Magnetic field in a toroid coil :

$$ \begin{equation*} B=\frac{\mu_{0} N I}{2 \pi r} \tag{1} \end{equation*} $$

Now,

$$ B=\frac{800 \times 4 \pi \times 10^{-7} \times 4000 \times 3}{2 \pi \times 20 \times 10^{-2}} 1 / 2 $$

$$ \begin{equation*} B=9.6 \mathrm{~T} \tag{1} \end{equation*} $$

It is observed that as Bismuth is diamagnetic substance with relative permeability less than 1 , it will have a tendency to move away from the stronger to weak part of external magnetic field making the core field less as compared to empty core field.

[AI Q. 4. (i) State Ampere’s circuital law, expressing it in the integral form.

(ii) Two long co-axial insulated solenoids, $S_{1}$ and $S_{2}$ of equal lengths are wound one over the other as shown in the figure. A steady current " $I$ " flows through the inner solenoid $S_{1}$ to the other end $B$, which is connected to the other solenoid $S_{2}$ through which the same current " $I$ " flows in the opposite direction so as to come out at end $A$. If $n_{1}$ and $n_{2}$ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point (i) inside on the axis and (ii) outside the combined system.

U [Delhi I, II, III 2014]

Ans. (i) Ampere’s circuital law : The lineintegral of the magnetic field around a closed path is $\mu_{0}$ times of total current enclosed by the path.

$$ \oint \vec{B} \cdot \overrightarrow{d l} \leftrightharpoons \mu_{0} I $$

(Award 1 mark if the student just writes the integral form of Ampere’s circuital law)

(ii) (a)

$$ B=\mu_{0} n I $$

Magnitude of net magnetic field inside the combined system on the axis,

$$ \begin{aligned} & B=B_{1}-B_{2} \ \Rightarrow \quad & B=\mu_{0}\left(n_{1}-n_{2}\right) I \end{aligned} $$

$$ \text { Also accept if the student writes } B=\mu_{0}\left(n_{2}-n_{1}\right) I $$

(b) Outside the combined system, the net magnetic field is zero.

[CBSE Marking Scheme 2014]