Consider a toroidal solenoid wound on a non-magnetic core, such as shown in Fig. 20. If the flux density is measured on the centre line of the toroid, the relationship between B and *H *is given. b-y the straight line *OA *in Fig. 21. Ifnow the space within an unmagnetised ferromagnetic material, the well known magnetisation curve *OBCDE, *is obtained. The magnetisation has many names being referred to as :

*B-H curve, the magnetic saturation curve, the virgin curve or simply the saturation curve. *

The difference in flux between the saturation curve and the air line, *OA *at any magnetising force,is due to the contribution of the magnetic material. This flux is known as the *intrinsic flux *and gives a true measure of magnetic properties than does the total flux, especially at very high m.m.fs, The point at which the intrinsic flux density curve becomes horizontal gives the intrinsic saturation.

The magnetisation curves can be determined by the following methods provided the materials are in the. form of a ring:

*(i) *By means of a ballistic galvanometer

*(ii) *By means of a fluxmeter.

The graphs in Fig. 22 show the relationship between the flux density *(B) *and the magnetic field obtained for different qualities ofiron. The data for mild steel, wrought iron and sheet steel are so similar that they can be represented by a common graph. *Stalloy *is an alloy of iron and silicon commonly used in the construction of transformers and A.C. machines.

**Example 6**. *An air cored toroidal coil has **1500 **turns and carries a current of **0.2 *A *The **length of **the magnetic circuit is **300 **mm and cross-sectional area of the coil is **800 **mm ^{2}. Calculate: *

*(i) **Magnetic field strength, *

*(ii) **Flux density, and *

*(iii) **Total flux within the coil . *

**Solution**. Number of turns, *N *= 1500

Current, *I *= 0.2 A

Length of the magnetic circuit, *1 *= 300 mm = 0.3 m

Cross-sectional area of the coil, *A *= 800 mm^{2} = 800 × 10^{-6} m^{2}

I. ** ****Magnetic field strength, H : **

**III.Total Flux within the coil, ɸ:**

ɸ = B × A

= 1.256 × 10^{-3} × 800 × 10^{-6} = **1.005 ****× ****10 ^{-6} **

**Wb.**

**(Ans.)**

**Example ****7**. A *coil of **300 **turns and of resistance *10 Ω *is wound uniformly over a steel ring of *mean circumference *30 cm and cross-sectional area *9 *cm ^{2}. It is connected to a supply at *

*20*

*V.*

*D.C. If*rollll’

*III’rlllc”ability of the ring is*

*1500,*

*find:*

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