Magnetic flux density or magnetic induction

(a)       The magnetic lines of force passing through unit normal area in a magnetic field is defined as magnetic induction.

(b)       B = Φ / A when A = 1m². then B = Φ

(c)        Direction of magnetic induction: The direction in the magnetic field in which if a current carrying conductor is placed then no force acts on it, is known as the direction of magnetic induction.

(d)       Magnetic induction is a vector quantity.

(e)       The magnetic induction due to a bar magnet

            (i)         In axial position B = 2KM / r³

            (ii)        In equatorial position B = KM / r³

                        Here K = m₀ / 4p = 10^–7 Weber / A.m

1.2       Magnetising field or intensity of magnetic field H

(a)       The ratio of magnetic induction produced in vacum (B^o) and magnetic permeability of vacuum is defined as magnetising field (H), i.e. H = B₀ / μ₀

(b)       The intensity of magnetic field due to a pole of strength m_p at a distance r from it is H = m_p/ μr²

(c)        Due to a small magnet H = M/r³ √1+3 cos² θ

Illustration 2:      1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m² is produced in it. The magnetising force generated in the material will be :

(A)      1.2 × 10^–3 A/m                (B)      2.6 × 10^–3 A/m

(C)      2.6 × 10^–4 A/m                (D)      2 × 10³ A/m

Solution:             (D) H = Ni / 2pr = ni = 10³ × 2 = 2 × 10³ A/m

Illustration 3:      In the above problem, the value of intensity of magnetisation in A/m will be :

                              (A)      7.96 × 10⁶                        (B)      7.96 × 10^–6

                              (C)      3.98 × 10³                        (D)      zero

Solution:             (A) B = m₀ (I + H)

                              I = B / m₀ – H = 10/4p × 10^–7 – 2 × 10³ = 7.96 × 10⁶ A/m

Illustration 4:      In the above question 115 the relative permeability of the material will be :

                              (A)      4.98 × 10³                        (B)      4.98 × 10^–3

                              (C)      2.98 × 10^–3                       (D)      3.98 × 10³

Solution: