Question bank Class 12 Physics Chapter 11. Magnetic materials
Standard:- 12th
Question Bank
Subject:- Physics
Chapter 11. Magnetic materials
MCQ’S (1 Mark Each)
1) The magnetic susceptibility is given by
2) The relation between relative permeability and magnetic susceptibility is given by
a) χ = μ + 1 b) χ = −μ −1 c) μ = 1 − 𝜒 d) μ = 1 + 𝜒
Ans : d) μ = 1 + 𝜒
3) If an electron of charge (-e) and mass 𝑚 revolves around the nucleus of an atom having orbital magnetic moment 𝑚,then angular momentum of electron is
4) If 𝑚 and L denote the orbital angular moment and the angular momentum of the electron due to its orbital motion, then the gyromagnetic ratio is given by
5) Relative permeability of iron 5500,then its magnetic susceptibility will be
a) 5500 × 10 b) 5501 c) 5499 d) 5500× 10
Ans : c) 5499 , hint : χ = μ − 1
6) What is magnetization of a bar magnet having length 6 cm and area of cross section
5 𝑐𝑚 ?
( 𝑚= 1)
𝑎)1.2 × 10 A/m b) 3.3 × 10 A/m c) 1.25 × 10 A/m d) 3.3 × 10 A/m
Ans :
7) A magnetic material of susceptibility 3 × 10, and magnetic intensity is 4 × 10
A𝑚 .Then The magnetization in the units of A𝑚 is
a) 12 × 10 b) 1.33 × 10 c) 0.75 × 10 d) 14 × 10
Ans : a) 12 × 10 , M = χ H
Very Short Answer (VSA) ( 1 MARK Each )
1. Give gyro magnetic ratio.
2. What is stated in term of Bohr magneton.
3. Define magnetization
4. What does the ratio of magnetization to magnetic intensity indicates?
5. The relative permeability of a medium is 0.075. What is its magnetic susceptibility ?
(Ans : 0.925 , hint : χ = μ − 1)
6. The moment of a magnet ( 15 cm× 2𝑐𝑚 × 1𝑐𝑚 ) is 1.2 A-𝑚.What is its intensity of magnetization? (Ans : 𝑀 = 4 × 10 A𝑚 )
7. The electron in hydrogen atom is moving with a speed of 2.5 × 10 m/s in an orbit of radius 0.5 𝐴°. What is the Magnetic moment of the revolving electron . (Ans : 𝑚 =
Short Answer I (SA1) ( 2 MARKS Each )
1. Show that the orbital magnetic dipole moment of a resolving electron is
2. Derive the quantity for Bohr magneton and also state its value.
3. Define magnetization. State its SI unit and dimensions.
4. Calculate the gyro magnetic ratio of electron.
(given e =1.6 × 10 C, 𝑚= 9.1 × 10 kg)
Ans: hint : gyromagnetic ratio =
= 8.8 × 10 C 𝑘𝑔
5.. An iron rod of area of cross-section 0.1𝑚 is subjected to a magnetizing field of 1000 A/m. Calculate the magnetic permeability of the iron rod. (χ for iron = 599 , μ = 4𝜋 × 10 SI unit )
Ans: 7.54× 10 , hint: μ = χ+1 , μ = μμ
6 A solenoid has core of a material with relative permeability 500 and its windings carry current of 1 A. The number of turns of the solenoid is 500 per metre . Calculate the magnetization of the material.
Ans: 2.495 × 10 A𝑚, H = Ni , χ = μ − 1, M= ΧH =(μ − 1)H
Short Answer II (SA2) ( 3 MARKS Each )
1) Define magnetic intensity. Explain magnetization of a material.
2) Obtain an expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.
3) Define gyromagnetic ratio.Find relation for Bohr magneton.
4) When a plate of magnetic material of size 10 𝑐𝑚 × 0.5 𝑐𝑚 × 0.2 𝑐𝑚 (length,breadth and thickness respectively ) is located in magnetizing field of 0.5 × 10 A𝑚 then magnetic moment of 5 A𝑚 is induced in it.Find out magnetic induction in the plate. ( Ans : M = 5× 10 𝐴𝑚, 𝐵 = 6.29 T)
5) A magnet of magnetic moment 3𝐴𝑚 weighs 75 g .The density of the material of the magnet is 7500 𝑘𝑔/𝑚 .what is the magnetization.?
(Ans : M= 3 × 10 A/m )
6) Find the relative permeability ,if the permeability of a metal is 0.1256 Tm𝐴
(Ans : 10 )
Long Answer ( LA) ( 4 marks Each)
1) Define magnetization. State its SI unit and dimensions. Derive the relation between magnetic field intensity(H) and magnetization(M) for a magnetic material placed in magnetic field.
2) Explain origin of magnetism in material, hence find magnetic moment of electron revolving around the nucleus of an atom.
3) An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10 m, with a speed of 2 × 10 m𝑠 .Find resultant orbital magnetic moment and angular momentum of electron.
(e = 1.6 × 10 C , m = 9.1 × 10 kg )
(Ans : 𝑚 = 8.48 × 10 𝐴𝑚 ,L = 9.646 × 10 𝑘𝑔. 𝑚/𝑠 )
