Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.26

Additional Exercise

# 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? g = 9.8 m s⁻².

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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.

Total number of turns, n = 3 x 300 = 900,

Length of the wire, 1 = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 x 10⁻

^{3}kgCurrent flowing through the wire, i = 6 A

Acceleration due to gravity, g = 9.8 m/s

^{2 }Magnetic field produced inside the solenoid,

B= (μ0nI)/LWhere, μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹

I = Current flowing through the windings of the solenoid

Magnetic force is given by the relation,

F = Bil= (μ0nI)/L x i

lAlso, the force on the wire is equal to the weight of the wire,

Therefore , mg=(μ0nI)/L x i

lI = (mgL)/(μ0nil) = (2.5 x 10⁻

^{3 }x 9.8 x 0.6 )/ ( 4π x 10⁻⁷ x 900 x 0.02 x 6)=108 A

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