Question: Q. 1. (i) State Ampere’s circuital law. Consider a long straight wire of a circular cross-section (radius $a$ ) carrying a steady current $I$.
(ii) The current $I$ is uniformly distributed across this cross-section. Using Ampere’s circuital law, find the magnetic field in the region $r<a$ and $r>a$.
R [SQP I 2017]
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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, $\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I$
where,
$B=$ magnetic field
$d l=$ 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 : $\quad \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I$
(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.
$$ \begin{array}{rlrl} & & \oint \vec{B} \cdot \overrightarrow{d l} & =\mu_{0}(n \cdot 2 \pi r l) \ & B \times 2 \pi r & =\mu_{0} n 2 \pi r I \end{array} $$
(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 $m=I A$
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.
$$ \text { Mathematically, } \oint \vec{B} \cdot \vec{d}=\mu_{0} I $$
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 $P$,
$$ d B= $$
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 $=I$.
$$ \begin{aligned} d B & =\frac{\mu_{0} n I d x a^{2}}{2\left(a^{2}+(x-x)^{2}\right)^{3 / 2}} \ & \because r \gg x, a \ & \therefore\left(a^{2}+(x-x)^{2}\right)^{3 / 2} \approx r^{3} \ \int_{0}^{B} d B & =\int_{-4 L} \frac{\mu_{0} n I d x a^{2}}{2 r^{3}} \ B & =\frac{\mu_{0} n I a^{2} l}{2 r^{3}} \ & =\frac{\mu_{1}}{4 \pi} \frac{2 n I l \pi a^{2}}{r^{3}}=\frac{\mu_{0}}{4 \pi} 2 \times \frac{\mathrm{NAI}}{r^{3}} \end{aligned} $$
$$ \begin{aligned} {[\because n \times l} & =N \ \pi a^{2} & =A] \end{aligned} $$
for a bar magnet.
$$ B=\frac{\mu}{4 \pi} \frac{2 m}{r^{3}} \text { (on avis) } $$
[Topper’s Answer 2015]