A uniform conducting wire of length $12 a$ and resistance $R$ is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side $a$, (ii) a square of sides $a$ and, (iii) a regular hexagon of sides $a$. The coil is connected to a voltage source $V_0$. Find the magnetic moment of the coils in each case.
We know that magnetic moment of the coils $m=n I A$.
Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.
(i) for an equilateral triangle of side $a$, $n=4$ as the total wire of length $=12 a$
Magnetic moment of the coils $m=n I A=4 I\left(\frac{\sqrt{3}}{4} a^2\right)$
$$\therefore \quad m=I a^2 \sqrt{3}$$
(ii) For a square of sides $a$,
$A=a^2$
$n=3$ as the total wire of length $=12 a$
Magnetic moment of the coils $m=n I A=3 I\left(a^2\right)=3 I \mathrm{a}^2$
(iii) For a regular hexagon of sides a,
$n=2$ as the total wire of length $=12 a$
Magnetic moment of the coils $m=n / A=21\left(\frac{6 \sqrt{3}}{4} a^2\right)$
$$m=3 \sqrt{3} a^2 I$$
$m$ is in a geometric series.
Consider a circular current-carrying loop of radius $R$ in the $x$-yplane with centre at origin. Consider the line integral
$$\Im(L)=\left|\int_{-L}^L \mathbf{B} \cdot \mathbf{d l}\right|$$
taken along $z$-axis.
(a) Show that $\Im(L)$ monotonically increases with $L$
(b) Use an appropriate amperian loop to show that $\Im(\infty)=\propto_0 I$. where $I$ is the current in the wire
(c) Verify directly the above result
(d) Suppose we replace the circular coil by a square coil of $\operatorname{sides} R$ carrying the same current $I$.
What can you say about $\Im(L)$ and $\Im(\infty)$ ?
(a) B (z) points in the same direction on $z$-axis and hence, $J(L)$ is a monotonically function of L.
Since, $B$ and $\mathbf{d l}$ along the same direction, therefore $\mathbf{B} . \mathbf{d l}=\mathbf{B} \cdot \mathrm{dl}$ as $\cos 0=1$
(b) $J(L)+$ contribution from large distance on contour $C=\propto_0 I$
$$\begin{aligned} & \therefore \quad \text { as } L \rightarrow \infty \\ & \text { Contribution from large distance } \rightarrow 0\left(\text { as B } \propto 1 / r^3\right) \\ & \qquad J(\infty)-\alpha_0 I \end{aligned}$$
(c) The magnetic field due to circular current-carrying loop of radius $R$ in the $x$-y plane with centre at origin at any point lying at a distance of from origin.
$$ \begin{aligned} B_z & =\frac{\alpha_0 I R^2}{2\left(z^2+R^2\right)^{3 / 2}} \\ \int_{-\infty}^{\infty} B_z d z & =\int_{-\infty}^{\infty} \frac{\alpha_0 I R^2}{2\left(z^2+R^2\right)^{3 / 2}} d z \\ \text{Put}\quad z & =R \tan \theta_1 \\ \Rightarrow\quad d z & =R \sec ^2 \theta d \theta \\ \therefore\quad\int_{-\infty}^{\infty} B_z d z & =\frac{\propto_0 I}{2} \int_{-\pi / 2}^{\pi / 2} \cos \theta d \theta=\propto_0 I \end{aligned}$$
(d) $B(z)_{\text {square }}
$$\therefore \quad \Im(L)_{\text {square }}<\Im(L)_{\text {circular coil }}$$
But by using arguments as in (b)
$$\Im(\infty)_{\text {square }}=\Im(\infty)_{\text {circular }}$$
A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure $10 \mathrm{~mA}, 100 \mathrm{~mA}$ and 1 mA using a galvanometer of resistance $10 \Omega$ and that produces maximum deflection for current of 1 mA . Find $S_1, S_2$ and $S_3$ that have to be used.
$$\begin{aligned} I_G \cdot G & =\left(I_1-I_G\right)\left(S_1+S_2+S_3\right) \text { for } I_1=10 \mathrm{~mA} \\ I_G\left(G+S_1\right) & =\left(I_2-I_G\right)\left(S_2+S_3\right) \text { for } I_2=100 \mathrm{~mA} \\ \text{and}\quad I_G\left(G+S_1+S_2\right) & =\left(I_3-I_G\right)\left(S_3\right) \text { for } I_3=1 \mathrm{~A} \\ \text{gives}\quad S_1 & =1 \mathrm{~W}, S_2=0.1 \mathrm{~W} \\ \text{and}\quad S_3 & =0.01 \mathrm{~W} \end{aligned}$$
Five long wires $A, B, C, D$ and $E$, each carrying current $I$ are arranged to form edges of a pentagonal prism as shown in figure. Each carries current out of the plane of paper.
(a) What will be magnetic induction at a point on the axis 0 ? Axis is at a distance $R$ from each wire.
(b) What will be the field if current in one of the wires (say $A$ ) is switched off?
(c) What if current in one of the wire (say $A$ ) is reversed?
(a) Suppose the five wires $A, B, C, D$ and $E$ be perpendicular to the plane of paper at locations as shown in figure.
Thus, magnetic field induction due to five wires will be represented by various sides of a closed pentagon in one order, lying in the plane of paper. So, its value is zero.
(b) Since, the vector sum of magnetic field produced by each wire at $O$ is equal to 0 . Therefore, magnetic induction produced by one current carrying wire is equal in magnitude of resultant of four wires and opposite in direction.
Therefore, the field if current in one of the wires (say $A$ ) is switched off is $\frac{\alpha_0}{2 \pi} \frac{i}{R}$ perpendicular to $A O$ towards left.
(c) If current in wire $A$ is reversed, then total magnetic field induction at $O$ $=$ Magnetic field induction due to wire $A+$ magnetic field induction due to wires $B, C, D$ and $E$
$$=\frac{\propto_0}{4 \pi R} \frac{2 I}{R}$$
(acting perpendicular to $A O$ towards left) $+\frac{\propto_0}{\pi} \frac{2 I}{R}$ (acting perpendicular $A O$ towards left) $=\frac{\propto_0 I}{\pi R}$ acting perpendicular $A O$ towards left.