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Moving Charges And Magnetism

Question
CBSEENPH12037677

An iron core is inserted into a solenoid 0.5 m long with 400 turns per unit length. The area of cross-section of the solenoid is 0.001 m2. (a) Find the permeability of the core when a current of 5 A flows through the solenoid winding. Under these conditions, the magnetic flux through the cross-section of the solenoid is 1.6 x 10–3 Wb. (b) Find the inductance of the solenoid under these conditions.

Solution
Given,
Length of the solenoid, l = 0.5 m
Number of turns per unit length, n = 400
Area of cross- section of the solenoid, A= 0.001 m2 

The magnetic induction on the axis of the solenoid is given by 

            B = μμ0nIB = μμ04π 4πni                  ....(1) 

where,
μ is the permeability of the medium,
n the number of turns per unit length and,
i is the current passing through the coil. 

(a) The normal to the area is along the direction of the field.

Therefore, 

Magnetic flux, ϕ = BA, 

Given,            Flux, ϕ = 1.6 × 10-3 Wb,Area,  A = 0.001 = 10-3 m2 

Therefore,       B = ϕA     = 1.6 × 10-310-3     = 1.6 Wb m-2  


Since,              No. of turns, n = 400, Current,  i = 5 A  and,μ04π = 10-7 Hm-1 

We have form equation (1),

        1.6 = μ × 10-7 × 4π × 400 × 5 

which gives,    
     μ = 1.6 × 1074π × 5 × 400 = 636.7  637 

(b) Total number of turns in the solenoid is given by

 N = n × l = 400 × 0.5 = 200 


Total flux through the solenoid is,

N ϕ = 200 × 1.6 × 10-3 = 0.32 Wb

Self inductance,

L = i = 0.325 = 0.064 H = 64 mH 

Alternatively,

               
L = μμ0N2Al = μμ0n2lA       
                            = 637 × (4π×10-7) × 400 × 400 × 0.5 × 10-3= 6.4 × 10-2H = 64 mH