Question

Question 4.26: 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

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-172

Answer 1

magnetic field in the solenoid is $B=\mu_0nI$
where n is the number of turns per unit length
I is the current passing through solenoid.

here, n = total turns /total length of solenoid
= 300 × 3/(60 × 10^-2) = 1500 turns/m
B = 4π × 10^-7 × 1500 × I [$\mu_0$=4π×10^-7]
B = 18.84I × 10^-4 T
A current carrying wire is suspended inside solenoid and a current I = 6A is flowing on it .
to balanced the weight of wire by the magnetic force
BIL= mg
18.84I × 10^-4 × 6 × 2 × 10^-2 = 2.5 × 10^-3 × 9.8
I = 108.4A