Solution:

Ans. (i) Ampere’s circuital law states that the line integral

of magnetic field around a closed path is ${\mu }_0$ times o

total current enclosed by the path, $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}I$

where,

$B=$ magnetic field

$dl=$ infinitesimal segment of the path

${\mu }_0=$ permeability of free space.

$I=$ enclosed electric current by the path

Magnefic field at a point will not depend on the

shape of Amperian loop and will remain same at every point on the loop.

(ii) Try yourself, Similar to Q. 6 Short Answer Type-II

AI] Q. 2. (i) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius , having ’ $n$ ’ turns per unit length and carrying a steady current $I$.

(ii) An observer to the left of a solenoid of $N$ turns each of cross section area ’ $A$ ’ observes that a steady current $I$ in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment $m=$ NIA.

R U [Delhi I, II, III 2015]

Ans.(i) Line integral of magnetic field over a closed loop is equal to the ${\mu }_0$ times the total current passing through the surface enclosed by the loop.

Alternatively : $\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}I$

(a)

(b)

Let the current flowing through each turn of the toroid be $I$. The total number of turns equals $n$. $(2\pi r)$ where, $n$ is the number of turns per unit length.

Applying Ampere’s circuital law, for the Amperian loop, for interior points.

$$\begin{array}{rlrlll}& & \oint \overrightarrow{B}\cdot \overrightarrow{dl}& ={\mu }_0(n\cdot 2\pi rl)& B\times 2\pi r& ={\mu }_{0}n2\pi rI\end{array}$$

(ii)

The solenoid contains $N$ loops, each carrying a current $I$. Therefore, each loop acts as a magnetic dipole. The magnetic moment for a current $I$, flowing in loop of area (vector) $A$ is given by $m=IA$

The magnetic moments of all loops are aligned along the same direction. Hence, net magnetic moment equals to NIA.

Ans. (i) Ampere’s circuital law deters that the cloned-losp intigral of the

the current passing through the loop.

$$\text{ Mathematically, }\oint \overrightarrow{B}\cdot \overrightarrow{d}={\mu }_{0}I$$

Total incoming current $=$ Total out going lurrent

$\therefore I_{\text{rut }}=0$

For loop II,

Applying ampere’s circuital law.

The magnetic files lines will be, as shown $\therefore$ Field line best in from North.

heme end B is Noria end $A$ is South

Consider a solenoid as shown Now tate an elementary lop of width che.

for this loop, yield at $P$,

$$dB=$$

then cement in each torn = I. then of no. of turn per vent length =n current in pathourn = cement in each torn = $I$. . of turn per vent length cement in paction =

then no. of torn per unit length cement is each torn $=I$.

$$\begin{array}{rlrlrlr}dB& =\frac{{\mu }_{0}nIdx{a}^2}{2{(a^2+(x-x{)}^2)}^{3/2}}& \because r\gg x,a& \therefore {(a^2+(x-x{)}^2)}^{3/2}\approx r^3{\int }_0^{B}dB& ={\int }_{-4L}\frac{{\mu }_{0}nIdx{a}^2}{2r^3}B& =\frac{{\mu }_{0}nI{a}^{2}l}{2r^3}& =\frac{{\mu }_1}{4\pi }\frac{2nIl\pi a^2}{r^3}=\frac{{\mu }_0}{4\pi }2\times \frac{\mathrm{NAI}}{r^3}\end{array}$$

$$\begin{array}{rlr}[\because n\times l& =N\pi a^2& =A]\end{array}$$

for a bar magnet.

$$B=\frac{\mu }{4\pi }\frac{2m}{r^3}\text{ (on avis) }$$

[Topper’s Answer 2015]