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Electric Charges And Fields

Question
CBSEENPH12037891

A toroidal solenoid with an air core has an average radius of 0.15 m, area of cross-section 12 x 10–4 m2 and 1200 turns. Obtain the self-inductance of the toroid. Ignore field variation across the cross-section of the toroid.

Solution

(a) Given, a toroidal solenoid. 
Average radius, r = 0.15 m
Area of cross-section, A = 12 x 10–4 m
Number of turns in the coil, n = 1200
Magnetic field is given by,
         
        B = μ0n1 I = μ0N1Il = μ0N1I2πr 

Total magnetic flux, ϕB = N1BA = μ0N12IA2πr

But,              ϕB = LI
                        L = μ0N12A2πr 

L = 4π×10-7×1200×1200×12×10-42π×0.15H
  = 2.3 × 10-3H= 2.3 mH
 

(b) Emf induced, E = ddtϕ2, 
where, Φ2 is the total magnetic flux linked with the second coil. 

Thus,

     E = ddtN2BA = ddtN2μ0N1I2πrA

     E = μ0N1N2A2πrdIdt 

     E = 4π×10-7×1200×300×12×10-4×22π×0.15×0.05V

         = 0.023 V