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NCERT · Class 12 · Physics

Chapter 5: Magnetism and Matter

Practice each question, then open the answer or worked solution.

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Questions

25 questions
Q5.21Written answer
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10–2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
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Worked solution
The magnitude of one of the magnetic fields, B1 = 1.2 × 10–2 T
Let the magnitude of the other field is B2
The angle between the field, θ = 60°
At stable equilibrium, the angle between the dipole and the field B1, θ1 = 15°
Angle between the dipole and the field B2, θ2 =θ – θ1 = 45°
The torque due to the field B1 = Torque due to the field B2
MB1 sin θ1 = MB2 sin θ2
Here, M is the magnetic moment of the dipole
B2  = MB1 sin θ1/M sin θ2
= (1.2 × 10–2 ) x sin 150/sin 450 = 4.39 x 10-3 T
The magnetic field due to the other magnetic field is 4.39 x 10-3 T
Q5.22Written answer
Q 5. 22) A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me= 9.11 × 10–31 kg). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of the earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
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Worked solution
Energy of the electron beam, E = 18 keV = 18 x 103 eV = 18 x 103 x 1.6 x 10-19 J
Magnetic field, B = 0.04 G
Mass of the electron, me = 9.11 × 10–31 kg
Distance to which the beam travels, d = 30 cm = 0.3 m
The kinetic energy of the electron beam is
E = (1/2) mv2
\(\begin{array}{l}v=\sqrt{\frac{2\times 18\times 10^{3}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}= 0.795\times 10^{8} m/s\end{array} \)
The electron beam deflects along the circular path of radius r.
The centripetal force is balanced by force due to the magnetic field
Bev = mv2/r
r = mv/Be
= (9.11 x 10-31 x 0.795 x 108)/(0.04 x 10-4 x 1.6 x 10-19)
= (7.24 x 10-23)/(0.064 x 10-23)
= 113.125
Let the up and down deflection of the beam be x = r (1 – cosθ)
here, θ is the angle of deflection
sin θ = d/r
= 0.3/113.12 = 0.0026
θ = sin -1 (0.0026)= 0.14890
x = r (1 – cosθ) = 113.12 (1 – cos 0.14890)
= 113.12 (1- 0.999)= 113.12 x 0.01 = 1.13 mm
Q5.23Written answer
Q 5. 23) A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10–23 J T–1. The sample is placed under a homogeneous magnetic field of 0.64 T and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
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Worked solution
Number of diatomic dipoles =  2.0 × 1024
Dipole moment of each dipole, M’= 1.5 × 10–23 J T–1
Magnetic field strength, B1= 0.64 T
Cooled to a temperature, T1 = 4.2 K
Total dipole moment of the sample = n x M’ = 2.0 × 1024  x1.5 × 10–23 = 30
Degree of magnetic saturation = 15 %
Therefore, M1 = (15/100) x 30 = 4.5 J/T
Magnetic field strength, B2 = 0.98 T
Temperature, T2 = 2.8 K
The ratio of magnetic dipole from Curie temperature,
\(\begin{array}{l}\frac{M_{2}}{M_{1}}=\frac{B_{2}}{B_{1}}\times \frac{T_{1}}{T_{2}}\end{array} \)
\(\begin{array}{l}M_{2}=M_{1}\frac{B_{2}}{B_{1}}\times \frac{T_{1}}{T_{2}}\end{array} \)
\(\begin{array}{l}M_{2}=4.5\frac{0.98}{0.64}\times \frac{4.2}{2.8}\end{array} \)
= 18.52/1.79 = 10.34 J/T
Q5.24Written answer
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
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Worked solution
Mean radius of the Rowland ring = 15 cm
Number of turns = 3500
Relative permeability of the core, μr = 800
Magnetising current, I = 1.2 A
Magnetic field at the core, 
\(\begin{array}{l}B = \frac{\mu _{r}\mu _{0}IN}{2\pi r}\end{array} \)
\(\begin{array}{l}B = \frac{800\times 4\pi \times 10^{-7}\times 1.2\times 3500}{2\pi \times 0.15}= 4.48 T\end{array} \)
The magnetic field in the core is 4.48 T
Q5.25Written answer
The magnetic moment vectors µs and µl associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron, are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:
µs= –(e/m) S,
µl= –(e/2m)l
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
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Worked solution
Of the two, the relation µl = – (e/2m)l is in accordance with classical physics. It follows easily from the definitions of µl and l:
µl =  IA =  (e/T) πr2 ——(1)
l = mvr = m (2πr2/T) ——(2)
where r is the radius of the circular orbit which the electron of mass m and charge (–e) completes in time T. Dividing (1) by (2).
Clearly, µI/l = [ (e/T) πr2  ]/[m (2πr2/T) ] = –(e/2m)

Therefore, µI= (−e /2 m) l
Since the charge of the electron is negative (–e), it is easily seen that µI and l are antiparallel, both normal to the plane of the orbit.

Note µs/S, in contrast to µl /l, is e/m, i.e. twice the classically expected value. This latter result (verified experimentally) is an outstanding consequence of modern quantum theory and cannot be obtained classically.

Class 12 Physics Chapter 5 NCERT Solutions for Magnetism and Matter

Chapter 5, Magnetism and Matter of Class 12 Physics, is curated as per the current CBSE Syllabus 2023-24. This chapter of