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

The force on the wire due to the magnetic field inside the solenoid is given by F = BIL, where B is the magnetic field, I is the current in the wire, and L is the length of the wire. This force must balance the weight of the wire, which is given by W = mg, where m is the mass of the wire and g is the acceleration due to gravity.

So, we have BIL = mg.

The magnetic field inside a solenoid is given by B = μ0nI, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current in the solenoid.

Substituting this into the previous equation gives μ0nI^2L = mg.

We want to find the current I in the solenoid, so we rearrange this equation to give I = sqrt(mg / (μ0nL)).

The total number of turns in the solenoid is 3 layers * 300 turns/layer = 900 turns. The length of the solenoid is 60 cm = 0.6 m, so the number of turns per unit length is 900 turns / 0.6 m = 1500 turns/m.

Substituting these values into the equation gives I = sqrt((2.5 g * 9.8 m/s^2) / ((4π10^-7 T m/A) * 1500 turns/m * 2.0 cm)) = sqrt((0.0245 kgm/s^2) / ((4π10^-7 T m/A) * 1500 1/m * 0.02 m)) = sqrt((0.0245 N) / (1.8849610^-3 T*A)) = sqrt(13 A^2) = 3.6 A.

So, a current of 3.6 A in the windings of the solenoid can support the weight of the wire. The sense of circulation of the current should be such that the magnetic field inside the solenoid is upwards, to balance the weight of the wire.