(a) Let us consider a solenoid, whose $\;\;\text{ radius }\;=a$ $\;\;\text{ length }\;=21$ number of turns per unit length $=n$ current passing through the solenoid $=I$. Let us consider a small element $d x$ at distance $x$ from O.Magnetic field at $P$ (a point at a distance $r$ from $O$ ) is $dB = \frac{\mu_0 n\, dx\, I a^2}{2[(r-x)^2+a^2]^{3/2}}$Integrating $B = \int\limits_{-1}^{+1} \; \frac{\mu_0 n\, dx\, I a^2}{2[(r-x)^2+a^2]^{3/2}}$Putting $[(r-x)^2+a^2]^{3/2} = r^3$ $B = \frac{\mu_0 n I a^2 \times 21}{2 r^3}$ .......(i) Now, magnetic moment, $M = n \times 2 I \times I \times \pi a^2$ $\therefore n \times 2 I \times I = \frac{M}{\pi a^2}$ .......(ii)Putting in the eqn (i) $B = \frac{\mu_0 a^2 M}{2 r^3 \times \pi a^2} = \frac{\mu_0 M}{2 r^3 \times \pi} = \frac{\mu_0 2 M}{2 \pi r^3}$The same magnetic field is produced by a bar magnet.Hence, a current carrying solenoid behaves like a bar magnet.(b) Magnitude of magnetic dipole moment $M$ with the circular current loop having $n$ number of turns, carrying a current $I$ and of area $A$ is $|M| = n I A$. The direction of the magnetic dipole moment is perpendicular to the plane of the loop.Here $n=5$$I = 2\ \text{A}$$r = (\;\text{ radius }\;) = 7\ \text{cm} = 0.07\ \text{m}$$\therefore$ Area $= A = n r^2 = 3.14 \times (0.07)^2\ \text{m}^2$$\therefore$ Magnetic dipole moment$=5 \times 2 \times 3.14 \times (0.07)^2 = 0.154\ \text{Am}^2$ OR(a) $A B$ and $C D$ are two straight very long parallel conductors, carrying currents $I_1$ and $I_2$ , respectively, separated by a distance $a$ .The magnetic induction due to current $I_1$ in $A B$ at a distance $a$ is: $\;\; B_1 = \frac{\mu_0 I_1}{2 \pi a}$ ......(i) This magnetic field acts perpendicular to the plane of the paper and inwards. The conductor CD with current $I_2$ is situated in this magnetic field. Hence, force on a segment of length $l$ of $C D$ due to magnetic field $B_1$ is $F_1 = B_1 I_2 l$ .Substituting $B_1$ from eqn (i) $F_1 = \frac{\mu_0 I_2 I_1 l}{2 \pi a}$ ......(ii)By Fleming's Left Hand Rule, F acts towards left. Similarly, the magnetic induction due to current $I_2$ flowing in CD at a distance $a$ is $B_2 = \frac{\mu_0 I_2}{2 \pi a}$ ......(iii)This magnetic field acts perpendicular to the plane of the paper and outwards. The conductor $AB$ with current $I_1$ , is situated in this field. Hence, force on a segment of length $l$ of $A B$ due to magnetic field $B_2$ is $F_2 = B_2 I_1 l$Substituting $B_2$ from eqn (iii), $F_2 = \frac{\mu_0 I_2 I_1 l}{2 \pi \Omega}$ ........(iv) By Fleming's Left Hand Rule, this force acts towards right. These two forces given in equations (ii) and (iv) attract each other. Hence, two parallel wires carrying currents in the samedirection attract each other and if they carry currents in the opposite direction, repel each other.Definition of 1 Ampere:The force between two parallel wires carrying currents on a segment of length $l$ is $F = \frac{\mu_0 I_2 I_1 l}{2 \pi a}$Force per unit length is $\frac{F}{l} = \frac{\mu_0 I_2 I_1}{2 \pi a}$ If $I_1 = I_2 = 1\ \text{A}$ and ' $a$ ' $= 1\ \text{m}$ Then, $\frac{F}{l} = \frac{\mu_0 I_2 I_1}{2 \pi a} = \frac{4 \pi \times 10^{-7}}{2 \pi}$ $= 2 \times 10^{-7}\ \text{Nm}^{-1}$ So, $1\ \text{A}$ is defined as that constant current which when flowing through two parallel infinitely long straight conductors of negligible cross section and placed in air or vacuum at a distance of one meter apart, experience a force of $2 \times 10^{-7}\ \text{Nm}^{-1}$. (b) Dipole moment of the magnet $= M = 3\ \text{Am}^2$ $F =$ force applied at a distance $10\ \text{cm}$ from the centreIt is now in equilibrium at an angle $= \theta = 30^{\circ}$External magnetic field strength $= B = 0.25\ \text{T}$The magnet will be at rest when the total torque acting on it is 0 .It means that the torque due to applied force $F$ is equal to the torque due to magnetic force.Torque due to applied force $= F \times 10$Torque due to magnetic force $= M B \sin\; \theta$ $= 3 \times 0.25 \times \sin\; 30^{\circ}$Since, torque due to applied force $F =$ torque due to magnetic force, so $F \times 10 = 3 \times 0.25 \times \sin\; 30^{\circ}\;$ $F = \frac{1}{10} \times 3 \times 0.25 \times \sin\; 30^{\circ}\;$ $= \frac{1}{10} \times 3 \times 0.25 \times 0.5$ $\; = 0.0375\ \text{N}$If $F$ is withdrawn, the magnet will go back to its original position.