A solenoid 60 cm long and of radius 4 cm has 3 layer of windings 300 turns each. A 2.3 cm long wire of mass 2.5g lies inside the solenoide 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 6A 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?

Explanation

If I is the current in the windings of the solenoid and is the number of turns per unit length, then magnetic field inside the solenoid is given by:
` B = mu_(0)n I`
Here, ` n = "Total number of turns"/"Length of solenoid"`
` = (300 xx 3)/(60 xx 10^(-2)) = 1,500 ` turns/m
` :. " " B = 4pi xx 10^(-7) xx 1,500 xx I`
` = 6pi xx 10^(-4) xx I T `
The direction of this field is along the axis of solenoid. Now, a wire is placed perpendicular to the axis of solenoid.
`:. ` Magnetic force on the wire will be
` F = I.l B sin 90^(@)`
Here, ` I. = 6 A, l = 2 cm 2 xx 10^(-2) m`
`rArr" " F = 6 xx 2 xx 10^(-2) xx 6pi xx 10^(-4) xx I`
` = 72 pi xx 10^(-6) xx I N`
To support the weight of the wire, this force should be equal and opposite to the weight of the wire.
`rArr " " F = mg`
` rArr " " 72 pi xx 10^(-6) xx I = 2.5 xx 10^(-3) xx 9.8`
` rArr " " I = (2.5 xx 10^(-3) xx 9.8)/(72 pi xx 10^(-6)) A = 108.3 A`