A bar magnet is a rectangular or cylindrical object made of ferromagnetic materials like iron and steel. If a bar magnet is placed on a sheet of glass with some iron filings sprinkled around it, the filings form a certain pattern around the bar. This pattern shows that just like the positive and negative charge of an electric dipole, the magnet has two poles. The pattern formed by the filings gives the magnetic field lines of the bar magnet. These magnetic field lines are used to find the direction of the magnetic field. Following are a few important properties of magnetic field lines:

• The magnetic field lines of a magnet form continuous loops unlike the electric dipole, whose field lines start at the positive end and end at the negative end or diverge to infinity.
• At a given point, the tangent to the field line represents the direction of the net magnetic field at that particular point.
• Greater the number of field lines per unit area, greater the strength of the magnetic field.
• Magnetic field lines never intersect.

A solenoid is a long coil of insulated copper wire arranged in circular loops that generates a magnetic field when current is made to pass through it. A solenoid and a bar magnet have similar magnetic fields. When the south pole of a bar magnet is brought close to the part of the solenoid that is connected to the positive terminal of a cell, the bar magnet and the solenoid repel each other. As opposite poles attract each other, the end connected to the positive terminal of the cell acts as the north pole of the solenoid while the other end acts as the south pole.

Gauss’ Law of Magnetism states that the flux of a magnetic field through a closed surface is always zero. Here, the area vector is pointing out from the surface. Since magnetic field lines are continuous loops, all closed surfaces have the same number of magnetic field lines going in and coming out. Hence, the net magnetic flux through a closed surface is zero.

Net flux ϕ=∫B.dA

Gauss’s law of magnetism states that through any closed surface, the flux of B will always zero.

The strength of the earth’s magnetic field is not uniform across the earth’s surface. Its value is of the order, 10–5 T.

The earth’s magnetic field extends millions of kilometres into outer space, and the magnetic field lines are similar to that of an imaginary magnetic dipole at the centre of the earth. The axis of the dipole is not the same as the earth’s axis of rotation. The earth’s magnetic south pole is near the North Pole, while the magnetic north pole is near the South Pole. This is due to opposite poles attracting each other. The Earth’s magnetic field may be large in size but is actually weak in terms of field strength. Following are some of the factors that affect the magnitude and direction of the earth’s magnetic field:

• Magnetic declination: The angle between the actual north pole and the magnetic north pole is the magnetic declination. On the horizontal plane, the actual north pole is never the same and changes according to the position on the earth’s surface and time.
• Magnetic inclination: Also called the angle of dip, the magnetic inclination is the angle between the total magnetic field of the earth and the horizontal plane on the earth’s surface. The angle of dip is 0° at the magnetic equator and 90° at the magnetic poles.

A circulating electron in an atom has a magnetic moment. In a dense material, these moments add up vectorially and result in an overall magnetic moment greater than zero. Magnetisation ‘M’ of a magnet is equal to its net magnetic moment per unit volume. The formula for magnetisation is given by:

M = mnet/V

Now consider a solenoid with ‘n’ turns per unit length and the current passing through it as ‘I’. The magnetic field in the interior of the solenoid is then given by:

B0 = μonI

If the solenoid’s interior is filled with a non-zero magnetisation material , the field inside the solenoid will be stronger. The net magnetic field ‘B’ inside the solenoid is given by:

B = B0 + Bm,  where Bm is the field generated by the core material. Bm is proportional to the magnetisation of the material, M.

Mathematically, Bm = μoM, where, µ0 is the constant of permeability of a vacuum.

Now consider the magnetic intensity of a material, which is given by:

This equation shows that the total magnetic field is given by:

B = μo(H + M)

Here, the magnetic field produced by external factors such as the current in the solenoid is given as ‘H’ while the magnetic field produced by the nature of the core is given as ‘M’. ‘M’ is dependent on external factors and is given by:

M = χH, where χ is the magnetic susceptibility of the material.

, where µr is the relative magnetic permeability of a material.

Magnetic permeability is given by:

μ = μ0μr = μ0(1 + χ)

Materials can be classified as diamagnetic, paramagnetic or ferromagnetic.

• A material is diamagnetic if its susceptibility value, χ, is negative. Diamagnetic materials move from the strong section to the weak section of the external magnetic field. A magnet would repel diamagnetic materials.
• A material is paramagnetic if its susceptibility value, χ, is positive and small. Paramagnetic materials are magnetised in an external magnetic field, but the effect is very weak. In an external magnetic field, these materials move from a weak point to a section with a strong magnetic field. A permanent magnet will have a weak pull on paramagnetic materials.
• A material is ferromagnetic if its susceptibility value, χ, is large and positive. Ferromagnetic materials are strongly magnetised when placed in an external magnetic field. These materials move from a weak part to a section with a strong magnetic field. A magnet will have a strong pull on ferromagnetic materials.

1 . Why should I refer to MSVgo's NCERT Solutions for Class 12 Physics Chapter 5?

MSVgo NCERT Solutions for Class 12 Physics Chapter 5 is reliable because the solutions are detailed and have been prepared by experts. Using the MSVgo NCERT Solutions will be helpful while preparing for exams as there are several questions and exercises available for rigorous practice. Watching MSVgo videos on the chapter enhances learning and boosts confidence in the topic. The subject matter of the chapter is extensively covered, ensuring that no important topic is left out.

2 . Is the NCERT Solutions for Class 12 Physics Chapter 5 the best reference guide for the students?

Yes, the NCERT Solutions for Class 12 Physics Chapter 5 is a useful reference guide for students as all the necessary definitions, terms, explanations, solved questions, and more are included. Students are encouraged to refer to the NCERT solutions to know what to expect in the board exams, and prepare for it well.

3 . Discuss the pattern of the questions from Chapter 5 Physics Class 12.

There are six one-mark questions, fourteen two-mark questions, fifteen three-mark questions, and fifteen five-mark questions.