Question: Q. 3. A toroidal solenoid of mean radius $20 \mathrm{~cm}$ has 4000 turns of wire wound on a ferromagnetic core of relative permeability 800 . Calculate the magnetic field in the core for a current of $3 \mathrm{~A}$ passing through the coil. How does the field change, when this core is replaced by a core of Bismuth?
U] [O.D. Comptt. III 2017]
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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]