Title 5. MAGNETISM AND MATTER.pdf ✅

Revision Notes Magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis  Magnetic dipole moment of bar magnet: M = m × 2l = 2ml (where m = pole strength, 2l = magnetic length)  Direction of magnetic dipole moment is from South pole to North pole. Point lies on axial line of bar magnet:  Magnetic field at point P due to: north pole of magnet (N) B = + − μ π 0 2 4 ( ) ( ) m d l KEY TERMS Magnetic length: It is the distance between the two poles of a magnetic dipole. Magnetic dipole moment: It is the product of pole strength and separation between two poles. It is denoted by M. Magnetic intensity: It is the magnetic moment per unit volume. Magnetic dipole: A system of two equal and opposite magnetic poles separated by small distance. south pole of magnet (S) B m d l = − + μ π 0 2 4 ( ) ( ) Hence, resultant magnetic field at point P when 2l << d: B M = μ π 0 3 2 4 d where, M = m × 2l. Point lies on equatorial line of bar magnet  Magnetic field at point P due to: North pole of magnet (direction N–P) B m d l = + + μ π 0 2 2 4 ( ) ( ) South pole of magnet (direction P–S) B m d l = − + μ π 0 2 2 4 ( ) Hence, resultant at point P when 2l << d: B M d = μ π 0 3 4 where, M = m × 2l Torque on a magnetic dipole (bar magnet) in a uniform magnetic field [Board 2023]  A bar magnet with length 2l and pole strength m in a uniform magnetic field B at an angle q with force mB acting on North and South pole results a couple where torque t due to the couple is t = force × perpendicular distance t = F × NA Learning Objectives After going through this chapter, the students will be able to:  Describe a bar magnet and identify the features of a magnetic field due to the bar magnet.  Define a magnetic dipole and describe its interaction with the external magnetic field when placed parallel, anti-parallel or any other angle.  State and explain Gauss's law of magnetism.  Define magnetisation, magnetic intensity and magnetic susceptibility and differentiate different types of magnetic materials on the basis these properties. 5 CHAPTER MAGNETISM AND MATTER

Magnetism and Matter (b) The magnetic field B is given by B = μrμ0H = 400 × 4π × 10–7 (N/A2 ) × 2 × 103 (A/m) = 1.0T (c) Magnetisation is given by M = ( ) B − μ μ 0 0 H = ( ) μ μ μ μ r 0 0 H 0 − H = (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 core. Thus B = μrn(I + Im). Using I = 2A, B = 1T, we get Im = 794 A. Magnetism and Gauss’s Law: Consider a small vector area element ΔS of a closed surface S as given in the figure. ∆S q B n The magnetic flux through ∆S is defined as ∆fB = B⋅∆S, where B is the field at ∆S. We divide S into many small area elements and calculate the individual flux through each. Then, the net flux fB is, fB = ∆φB all ∑ = B S . ∆ all ∑ = 0 where ‘all’ stands for ‘all area elements ∆S. Example-2 (a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length, (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B.] Solution: (a) Regardless of the scenario, two magnets are obtained, each with a distinct north and south pole. (b) No force occurs if the magnetic field is uniform. The iron nail undergoes a non-uniform field due to the bar magnet, resulting in an induced magnetic moment. Consequently, the nail experiences both force and torque. The net force is attractive because the induced south pole is closer to the magnet's north pole than the induced north pole. (c) Not necessarily true; it holds only if the source of the magnetic field possesses a net non-zero magnetic moment. This condition is not met for a toroid or even a straight infinite conductor. (d) Attempt to bring the different ends of the bars closer. A repulsive force in certain situations indicates both bars are magnetized, while constant attraction suggests one is not magnetized. In a bar magnet, the magnetic field is strongest at the poles and weakest at the central region. To determine which bar is magnetized, pick up one (e.g., A) and lower its end onto both ends and the middle of the other bar (e.g., B). If there is no force experienced by A in the middle of B, then B is magnetized. If there's no change from end to middle of B, then A is magnetized. Magnetic field lines  Imaginary closed loops which continuously represent the direction of magnetic field at any point. Tangent at any point of these loops gives the direction of magnetic field at that point.  Concentration of field lines gives strength of magnetic field.  Two field lines can never intersect each other. If they would, there would be two tangents at the point of intersection, which means two directions of magnetic lines, which is impossible.  In a ‘uniform’ magnetic field, the field lines are parallel and equidistant.  Potential energy of a bar magnet in a magnetic field, U M = - B . Bar magnet as an equivalent solenoid  If a solenoid of length 2l, radius a with current I having n number of turns per unit length, then the magnetic moment of solenoid, M = NIA, B M d = μ π 0 3 2 4