(a) Let us consider a solenoid, whose
‌‌ radius ‌=a
‌‌ length ‌=21
number of turns per unit length =n
current passing through the solenoid =I.
Let us consider a small element dx at distance x from O.
Magnetic field at P (a point at a distance r from O ) is
dB=‌

µ0n‌dx‌I‌a2

2[(r−x)2+a2]3∕2

Integrating
B=

+1

∫

−1

‌

µ0n‌dx‌I‌a2

2[(r−x)2+a2]3∕2

Putting [(r−x)2+a2]3∕2=r3
B=‌

µ0nIa2×21

2r3

.......(i)
Now, magnetic moment, M=n×2I×I×πa2
∴n×2I×I=‌

M

πa2

.......(ii)
Putting in the eqn (i)
B=‌

µ0a2M

2r3xπa2

=‌

µ0M

2r3xπ

=‌

µ02M

2πr3

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|=n‌IA. The direction of the magnetic dipole moment is perpendicular to the plane of the loop.
Here n=5
I=2A
r=(‌ radius ‌)=7‌cm=0.07m
∴ Area =A=nr2=3.14×(0.07)2m2
∴ Magnetic dipole moment
=5×2×3.14×(0.07)2=0.154Am2
OR
(a) AB and CD are two straight very long parallel conductors, carrying currents I1 and I2 , respectively, separated by a distance a .
The magnetic induction due to current I1 in AB at a distance a is: ‌‌B1=‌

µ0I1

2πa

......(i)

This magnetic field acts perpendicular to the plane of the paper and inwards. The conductor CD with current I2 is situated in this magnetic field. Hence, force on a segment of length l of CD due to magnetic field B1 is F1=B1I2l .Substituting B1 from eqn (i)
F1=‌

µ0I2I1l

2πa

......(ii)
By Fleming's Left Hand Rule, F acts towards left. Similarly, the magnetic induction due to current I2 flowing in CD at a distance a is
B2=‌

µ0I2

2πa

......(iii)
This magnetic field acts perpendicular to the plane of the paper and outwards. The conductor AB with current I1 , is situated in this field. Hence, force on a segment of length l of AB due to magnetic field B2 is
F2=B2I1l
Substituting B2 from eqn (iii),
F2=‌

µ0I2I1l

2πΩ

........(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=‌

µ0I2I1l

2πa

Force per unit length is ‌

F

l

=‌

µ0I2I1

2πa

If I1=I2=1A and ' a ' =1m
Then, ‌

F

l

=‌

µ0I2I1

2πa

=‌

4π×10−7

2π

=2×10−7Nm−1
So, 1A 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 2×10−7Nm−1.
(b)
Dipole moment of the magnet =M=3Am2
F= force applied at a distance 10‌cm from the centre
It is now in equilibrium at an angle =θ=30∘
External magnetic field strength =B=0.25T
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×10
Torque due to magnetic force =MB‌s‌i‌n‌θ
=3×0.25×sin‌30∘
Since, torque due to applied force F= torque due to magnetic force, so
F×10=3×0.25×sin‌30∘‌
F=‌

1

10

×3×0.25×sin‌30∘‌ =‌

1

10

×3×0.25×0.5
‌=0.0375N
If F is withdrawn, the magnet will go back to its original position.