A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to the axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 ms^–2.

Given a solenoid.
Length of the solenoid,l = 60 cm = 0.60 m
Total number of turns, N = 3 x 300 = 900
Length of the wire,  l₁ = 2.0 cm = 0.02 m
mass of the wire lying inside the solenoid,m 2.5 g = 2.5 x 10^–3 kg
Current carried by the wire,I₁ = 6.0 A 

Let, current I be passed through the solenoid windings, then,
Magnetic field produced inside the solenoid due to current is

                      

Force acting on wire, $={\mathrm{I}}_{1 }{l}_1 \mathrm{B} = {\mathrm{I}}_{1 }{l}_{\mathit{1}}\frac{{\mathrm{\mu }}_0\mathrm{NI}}{l}$ 

The wire can be supported if the force on wire is equal to the weight of wire, i.e. 

               ${\mathrm{I}}_1{l}_1 \frac{{\mathrm{\mu }}_0\mathrm{NI}}{l}=\mathrm{mg}$

$\Rightarrow$            $$\begin{gathered} \mathrm{I} = \frac{\mathrm{mg}l}{{\mathrm{I}}_1{l}_1{\mathrm{\mu }}_0\mathrm{N}} \\ = \frac{2.5 \times {10}^{-3}\times 9.8\times 0.6}{\left(6.0\right)\times (0.02)\times (4\mathrm{\pi }\times {10}^{-7})}\times 900 \end{gathered}$$
     
                 $= 108.27 \mathrm{A}$

Given,
Current carried by wire, I = 0.40 A,
Radius of circular wire, r = 8.0 cm = $8\times {10}^{-2}\mathrm{m}$
Number of turns in the coil, $\mathrm{n} = 100$

Using the formula, we get the magnetic field at the centre of coil as

$\boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{nI}}{2\mathrm{r}}} = \frac{4\mathrm{\pi }\times {10}^{-7}\times 100\times 0.4}{2\times 8.0\times {10}^{-2}}\mathrm{T}$
               $= 3.1 \times {10}^{-4}\mathrm{T}.$
                        

Given, current, I= 90 A
distance ,r = 1.5 m
Thus,
the magnitude of magnetic field is given by Ampere's circuital law 

               $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 90}{2\mathrm{\pi }\times 1.5} \\ = 1.2 \times {10}^{-5} T \end{gathered}$$  

Applying the right-hand thumb rule, we find that the magnetic field at the observation point is directed towards south.

Given,
Current carried by conductor, I = 35 A
Distance, r = 20 cm = 0.2 m

Using Ampere's circuital law we get, 

 $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 35}{2\mathrm{\pi }\times 0.20}\mathrm{T} \\ = 3.5 \times {10}^{-5}\mathrm{T}. \end{gathered}$$