4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

5.4 MAGNETISATION AND MAGNETIC INTENSITY The earth abounds with a bewildering variety of elements and compounds. In addition, we have been synthesising new alloys, compounds and even elements. One would like to classify the magnetic properties of these substances. In the present section, we define and explain certain terms which will help us to carry out this exercise. We have seen that a circulating electron in an atom has a magnetic moment. In a bulk material, these moments add up vectorially and they can give a net magnetic moment which is non-zero. We define magnetisation M of a sample to be equal to its net magnetic moment per unit volume: mnet M = (5.7) V M is a vector with dimensions L–1 A and is measured in a units of A m–1. Consider a long solenoid of n turns per unit length and carrying a current I. The magnetic field in the interior of the solenoid was shown to be given by B0 = µ0 nI (5.8) If the interior of the solenoid is filled with a material with non-zero magnetisation, the field inside the solenoid will be greater than B0. The net B field in the interior of the solenoid may be expressed as B = B0 + Bm (5.9) 145 Reprint 2025-26 Physics where Bm is the field contributed by the material core. It turns out that this additional field Bm is proportional to the magnetisation M of the material and is expressed as Bm = µ0M (5.10) where µ0 is the same constant (permittivity of vacuum) that appears in Biot-Savart’s law. It is convenient to introduce another vector field H, called the magnetic intensity, which is defined by B H = – M (5.11) µ0 where H has the same dimensions as M and is measured in units of A m–1. Thus, the total magnetic field B is written as B = µ0 (H + M) (5.12) We repeat our defining procedure. We have partitioned the contribution to the total magnetic field inside the sample into two parts: one, due to external factors such as the current in the solenoid. This is represented by H. The other is due to the specific nature of the magnetic material, namely M. The latter quantity can be influenced by external factors. This influence is mathematically expressed as M = χH (5.13) where χ , a dimensionless quantity, is appropriately called the magnetic susceptibility. It is a measure of how a magnetic material responds to an external field. χ is small and positive for materials, which are called paramagnetic. It is small and negative for materials, which are termed diamagnetic. In the latter case M and H are opposite in direction. From Eqs. (5.12) and (5.13) we obtain, B = µ0(1 + χ)H (5.14) = µ0 µr H = µ H (5.15) where µr= 1 + χ, is a dimensionless quantity called the relative magnetic permeability of the substance. It is the analog of the dielectric constant in electrostatics. The magnetic permeability of the substance is µ and it has the same dimensions and units as µ0; µ = µ0µr = µ0 (1+χ). The three quantities χ, µr and µ are interrelated and only one of them is independent. Given one, the other two may be easily determined. Example 5.5 A solenoid has a core of a material with relative 5.5 permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current Im.146 EXAMPLE Reprint 2025-26 Magnetism and Matter Solution (a) The field H is dependent of the material of the core, and is H = nI = 1000 × 2.0 = 2 ×103 A/m. (b) The magnetic field B is given by B = µr µ0 H = 400 × 4π ×10–7 (N/A2) × 2 × 103 (A/m) = 1.0 T (c) Magnetisation is given by M = (B– µ0 H)/ µ0 = (µr µ0 H–µ0 H)/µ0 = (µr – 1)H = 399 × H ≅ 8 × 105 A/m (d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the EXAMPLE core. Thus B = µr n (I + IM). Using I = 2A, B = 1 T, we get IM = 794 A. 5.5