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NCERT Solutions Class 12 Physics Chapter 5 – Magnetism and Matter
5.1
Ans – According to insights shared in the above question; We get,
The overall magnetic field strength of B = 0.25T
The total torque present on the bar magnet is T = 4.5×10−2J
The angle between the desired bar magnet & the external magnetic field can be denoted as θ = 30∘
The torque, as opposed to magnetic moment (M), can be represented as the below equation:
5.2
Ans – (a) As we all know, it has been given that the moment of the bar magnet can constitute M = 0.32 J/T along with an external magnetic field known as B = 0.15T
Whenever an iron bar magnet is perfectly positioned across the field of magnetic attraction, it’s been presumed to have a position of stable equilibrium. As a result, the separation angle θ across the field of magnetic attraction along with the magnetic bar can be denoted as 0∘.
(b) As we all know, the above equation represents that the moment of iron bar magnet will be M = 0.32 J/T & the overall external magnetic field can be given as, B = 0.15T
The bar magnet is said to be operating in fragile equilibrium because it is oriented in opposition toward a field of magnetic attraction denoted by the equation θ = 180∘
The system’s energy potential therefore corresponds to −MBcos(θ)
5.3
Ans – As we all know from the above statement the total number of turns present in the solenoid element can be represented using n = 800.
The total area of the cross-section will be A = 2.5 × 10−4 m2
The existing current present in the solenoid is I = 3 A
Given that a field of magnetic attraction forms down the longitudinal direction of an electrical current-carrying solenoid—that is, along the path connecting both the north and south poles—it is comparable to an iron bar magnet. Calculating the magnetic moment that is caused by the specified current-carrying solenoid consists of the ones listed below:
M = nIA = 800 × 3 × 2.5 × 10−4 = 0.6 J/T
5.4
Ans – Considering the attraction of magnetic moment, M = 0.6 /T
The field’s magnetic power, B = 0.25 T
The solenoid’s plane of spins & the trajectory of the externally applied force create a turn angle which is θ = 30∘. Therefore, the torque exerted on the solenoid can be calculated as follows:
τ = MBsin(θ)
⇒ τ = 0.6 × 0.25 × sin(30∘) = 7.5 × 10−2 J
5.5
5.6
5.7
Class 12 Physics Chapter-5 Magnetism and Matter Overview
Have you ever thought about why a compass always points north or how Earth acts like a big magnet? Chapter 5 goes into these everyday wonders by looking at the properties of magnetic materials, magnetic field lines, Earth’s magnetism, and magnetic dipoles. Our Magnetism and Matter NCERT Solutions can help you understand ideas that can be hard to understand at first.
But a lot of students have trouble with this chapter since it mixes physics and geography. That’s why we use examples and analogies from the actual world to make it easier. For example, linking hysteresis and magnetic susceptibility to soft iron or steel helps you see what you’re learning instead of just memorizing it, which makes it more relevant.
This chapter has changed a lot since the 2025 NCERT curriculum. There has been greater interest in topics like the magnetic properties of materials and the Earth’s magnetic field. These new ideas are included in our Magnetism and Matter NCERT Solutions, along with answered questions that follow the current pattern.
Finally, this chapter is about making robust theoretical bases. Our solutions include short definitions, labeled graphics, and questions that are similar to those on the exam, so you’ll be ready for boards or competitive exams. You will get the clarity and confidence you need to solve even the hardest magnetism difficulties with practice.
FAQs – Magnetism and Matter Class 12 Chapter-5 NCERT
It’s easy to confuse the two. Magnetic intensity is due to the external field, while magnetization is due to the material. We explain both with comparisons.
Because Earth’s core generates a magnetic field! This concept is explained clearly in the syllabus with visual aids and solved questions in our solutions.
You’re not alone. Our solutions include stepwise breakdowns of the loop with proper labels and real-life analogies to help make sense of it.
We use a table format with examples, characteristics, and formulas side-by-side so it’s easier to recall during exams.