HC Verma Solutions Class 12 Chapter 15 Magnetic Properties Of Matter

HC Verma Solutions Class 12 Magnetic Properties of Matter Chapter 15 that are provided here for class 12 students and JEE aspirants will enable them to develop a strong base in tackling complex questions related to the chapter. The solutions will further help them to have a fresh problem-solving approach to exercises as well as gain a better picture of the concepts.

The HC Verma solutions are important resources for students who are planning to stay ahead in the competition. With these solutions, they will be able to significantly boost their preparation level and prepare well for the exams including competitive exams like JEE and NEET. The solutions for HC Verma book have been prepared by subject experts and are in the best interest of the students.

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Key Concepts In Magnetic Properties Of Matter

The key concepts covered in this chapter include;

  • Magnetization of Materials and Intensity of Magnetization
  • Paramagnetism, Ferromagnetism and Diamagnetism
  • Magnetic Intensity
  • Magnetic Susceptibility
  • Permeability
  • Curie’s Law
  • Hysteresis
  • Soft Iron and Steel

Class 12 Important Question In Chapter 15

1. Is it possible to make permanent magnets from paramagnetic materials?

2. It is often said that diamagnetism is present in all types of materials. Can you explain why some materials are classified as ferromagnetic or paramagnetic?

3. Why are soft iron used to make electromagnets?

(a) because of low retentivity and low coercive force (b) due to low retentivity and high coercive force (c) it has high retentivity and high coercive force (d) it possesses high retentivity and low coercive force  

4. What happens to the magnetic susceptibility when a ferromagnetic material goes through a hysteresis loop? State its value.

(a) negative (b) zero (c) it has a fixed value (d) may be infinity

5. For which material is the magnetic susceptibility generally negative?

(a) diamagnetic materials (b) ferromagnetic materials (c) paramagnetic materials (d) both paramagnetic and ferromagnetic materials

HC Verma Solutions Vol 2 Magnetic Properties of Matter Chapter 15

Question 1: The magnetic intensity H at the centre of a long solenoid carrying a current of 2.0 A, is found to be 1500 A m–1. Find the number of turns per centimeter of the solenoid.

Solution:

Current in the solenoid = I = 2A (given)

Magnetic intensity H at the centre = 1500 Am-1 (Given)

We know, Magnetic field produced by solenoid = B = μoni ..(1)

Where,

i = current in the solenoid

n = number of turns per unit length

μ0 = absolute permeability of vacuum

And, magnetizing field intensity (H) in the absence of any material :

H = B/μ0 …(2)

Where B is net magnetic field

From (1) and (2),

H = ni

1500 = n x 2

=> n = 750 turns/m or 7.50 turns/cm

Number of turns per cm of the solenoid is 7.5 turn/cm.

Question 2: A rod is inserted as the core in the current-carrying solenoid of the previous problem.

(a) What is the magnetic intensity H at the centre?

(b) If the magnetization I of the core is found to be 0.12 A m–1, find the susceptibility of the material of the rod.

(c) Is the material paramagnetic, diamagnetic or ferromagnetic?

Solution:

(a) Since the magnet and rod are long and we are interested in the magnetic intensity at the center, the end effects can be neglected. There is no effect of the rod on the magnetic intensity in the middle. So, magnetization intensity remains the same.

Therefore, magnetic intensity H at the centre = H = 1500 Am-1 (using from previous problem as given)

(b) Magnetization of the core = I = 0.12 Am-1

We know, I = χH

χ = susceptibility of material of the rod.

χ = 0.12/1500

=> χ = 8 x 10-5

(c) The material is paramagnetic.

Question 3: The magnetic field inside a long solenoid having 50 turns cm-1 is increased from 2.5 × 10–3 T to 2.5 T when an iron core of cross-sectional area 4 cm2 is inserted into it. Find

(a) the current in the solenoid,

(b) the magnetization I of the core and

(c) the pole strength developed in the core.

Solution:

Number of turns per cm = 50

Magnetic field inside solenoid with iron core = 2.5 T

Magnetic field inside without iron core = 2.5× 10-3 T

Cross-sectional area = 4cm2

Magnetic field inside the solenoid without iron core:

B = μ0ni

Or i = B/ μ0n

i = 2.5 x 10-3/[4πx10-7x5000] = 0.4 A

(b)

I = B/μ0 – H

B = 2.5 T (Known)

Where, I = Magnetization of the core and H=magnetization intensity

The difference between two magnetic fields strengths, say B1 and B2 is

HC Verma Vol2 Ch 15 Solution 3

(c) Intensity of magnetization:

l = M/V = m(2I)/A(2I) = m/A

=> m = lA

=> m = 2 x 102 x 4 x 10-4 = 800 A-m

Question 4: A bar magnet of length 1 cm and cross-sectional area 1.0 cm2 produces a magnetic field of 1.5 × 10–4 T at a point in end-on position at a distance 15 cm away from the centre.

(a) Find the magnetic moment M of the magnet.

(b) Find the magnetization I of the magnet.

(c) Find the magnetic field B at the centre of the magnet.

Solution:

(a) Magnetic field at a point in the axis of magnet:

B = μ0/4π x 2Md/(d2 – l2)2

Substituting given values,

1.5 x 10-4 = [3×10-8xM]/[5.01×10-4]

Or M = 2.5 A

(b) Intensity of magnetization:

I = M/V = 2.5/[10-4x10-2] = 2.5 x 106 A/m

(c) We know, H = M/4πld2

Total magnetic field intensity(H) is the sum of magnetic field intensities due to north pole and south pole, which are equal in magnitude.

=>H = M/4πld2 + M/4πld2 = M/2πld2

H = 2.5/[2×3.14×0.01x(0.15)2]

=>H = 2 x 884.6 Am-1

Now net magnetic at the centre B:

B = μ0 (H + I)

I = Intensity of magnetization

=>B = 4π x 10-7 x [(2 x 884.6) + (2.5×106)] = 3.14 T

Question 5: The susceptibility of annealed iron at saturation is 5500. Find the permeability of annealed iron at saturation.

Solution:

Susceptibility of iron at saturation = χ = 5500

We know,

μ = μ0(1 + χ)

μ = 4π x 10-7 (1 + 5500) = 6.9×10-3

μ = 6.9×10-3 Henry/m

Question 6: The magnetic field B and the magnetic intensity H in a material are found to be 1.6 T and 1000 A m-1 respectively. Calculate the relative permeability μ_r and the susceptibility χ of the material.

Solution:

Magnetic field in the material = 1.6 T

Magnetic intensity = H = 1000 Am-1

We know, μ = B/H and μr = μ/μo

=> μr = B/Hμo = 1.6/[1000×4πx10-7] = 1.3 x 103

Relation between χ and μr is

 

μr =1 + χ

or χ = 1.3 x 103 – 1 = 1299

or χ = 1.3 x 103 (approx.)

Question 7: The susceptibility of magnesium at 300 K is 1.2 × 10–5. At what temperature will the susceptibility increase to 1.8 × 10–5?

Solution:

Susceptibility at temperature T1, χ1 = 1.2 x 10-5

Susceptibility at temperature T2, χ2 = 1.8 x 10-5

According to Curie’s law, χ = C/T

We have, χ1/ χ2 = T1/ T2

=> [1.2 x 10-5]/ [1.8 x 10-5] = T1/ 300

=>T2 = 200 K

Question 8: Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magneton equals 9.27 × 10–24 A m2). The density of atoms in iron is 8.52 × 1028 atoms m–3. (a) Find the maximum magnetization I in a long cylinder of iron.

(b) Find the maximum magnetic field B on the axis inside the cylinder.

Solution:

We know, Intensity of magnetization = I = M/V …(1)

For 1m3 volume, the number of atoms will be 8.52 x 1028

So, total magnetic moment = M = 8.52× 1028× 2×9.27× 10-24Am-2

=>M = 1.58 x 106 Am2

(1)=> I = 1.58 x 106 Am-1 : maximum magnetization

Now,

The net magnetic field: B = μo(I + H)

Magnetizing field intensity (H) = 0, as we have to find the maximum magnetic field on the axis of cylinder.

So, B = μoI

B = 4π x 10-7 x 1.58 x 10-6 = 2 T (approx.)

Question 9: The coercive force for a certain permanent magnet is 4.0 × 104 A m–1. This magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetize it completely. Find the current.

Solution:

Coercive force for magnet = Magnetic intensity = H = 4.0× 104 Am-1

Number of turns per cm inside solenoid = 40

Number of turns per m inside the solenoid = 4000 turns/m

Magnetic field produced by solenoid at the centre :

B = μo nI

And magnetizing field intensity in the absence of any material:

H = B/ μo

=> H = nI (From above equations)

Or I = H/n = 4 x 104/4000 = 10 A, which is the required current.

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