A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each.
Question:

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 its 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 the 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? = 9.8 m s−2

Solution:

Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each.

$\therefore$ Total number of turns, $n=3 \times 300=900$

Length of the wire, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 × 10−3 kg

Current flowing through the wire, i = 6 A

Acceleration due to gravity, g = 9.8 m/s2

Magnetic field produced inside the solenoid, $B=\frac{\mu_{0} n I}{L}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

= Current flowing through the windings of the solenoid

Magnetic force is given by the relation,

$F=B i l$

$=\frac{\mu_{0} n I}{L} i l$

Also, the force on the wire is equal to the weight of the wire.

$\therefore m g=\frac{\mu_{0} n \text { Iil }}{L}$

$I=\frac{m \mathrm{~g} L}{\mu_{0} n i l}$

$=\frac{2.5 \times 10^{-3} \times 9.8 \times 0.6}{4 \pi \times 10^{-7} \times 900 \times 0.02 \times 6}=108 \mathrm{~A}$

Hence, the current flowing through the solenoid is 108 A.