# HC Verma ii Solutions for Class 12 Science Physics Chapter 15 - Magnetic Properties Of Matter

HC Verma ii Solutions for Class 12 Science Physics Chapter 15 Magnetic Properties Of Matter are provided here with simple step-by-step explanations. These solutions for Magnetic Properties Of Matter are extremely popular among class 12 Science students for Physics Magnetic Properties Of Matter Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma ii Book of class 12 Science Physics Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma ii Solutions. All HC Verma ii Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 285:

#### Question 1:

A paramagnetic material is placed in a magnetic field. Consider the following statements:

(A) If the magnetic field is increased, the magnetisation is increased.

(B) If the temperature is increased, the magnetisation is increased.

(a) A and B are true.

(b) A is true but B is false.

(c) B is true but A is false.

(d) A and B are false.

#### Answer:

(b) A is true but B is false.

if the magnetic field is increased, magnetisation of paramagnetic material placed in magnetic field is also increases. Hence, option (a) is correct.

Magnetization (*I*) is given by,

$\overrightarrow{I}=\frac{\overrightarrow{M}}{V}$

As the temperature is increased, magnetic moments of paramagnetic material becomes more randomly aligned due to incresed thermal motion. This leads decrease in the magnetization *I*. Hence, option (b) is incorrect.

#### Page No 285:

#### Question 2:

The property of diamagnetism is said to be present in all materials. Then, why are some materials paramagnetic or ferromagnetic?

#### Answer:

When a diamagnetic material is placed in magnetic field, dipole moment are induced in its atoms by the applied magnetic field. The direction of magnetic field due to induced dipole moment is opposite to the applied magnetic field therefore the resultant magnetic field is smaller than the applied magnetic field. This process is called diamagnetism. As this process takes place for all the material, therefore all the material exhibit diamagnetism. However, some material consists of atoms having some magnetic moment on their own (without applying magnetic field). As a result of it, when they are placed in magnetic field, they aligns their atomic dipole in the direction of applied magnetic field and hence their resultant magnetic field is more then the applied magnetic field and exhibit paramagnetism or ferromagnetism.

#### Page No 285:

#### Question 3:

Do permeability and relative permeability have the same dimensions?

#### Answer:

Magnetic permeability ($\mu $) is the ratio of magnetic flux density (*B*) to the magnetising field strength (*H*).

$\mu =\frac{B}{H}$

In CGS ( centimeter-gram-second) dimension of *B* and *H* is same. Hence, magnetic permeability is dimensionless. But in SI unit, dimension of *B* and *H* is not same. Thus, permeability is not dimensionless.

Relative permeability is defined as the ratio of magnetic permeabity of any medium to the permeability of the vaccum. Hence, it is dimensionless. Thus, permeability and relative permeability have the same dimensions in CGS system.

#### Page No 285:

#### Question 4:

A rod, when suspended in a magnetic field, stays in the east-west direction. Can we be sure that the field is in the east-west direction? Can it be in the north-south direction?

#### Answer:

No, it depends on the nature of rod. As we know that when the diamagnetic substance is suspended in a uniform field they set their longer axis right angles to the direction of magnetic field. So, if the material of the rod will be diamagnetic then it will stay in the east-west direction in perpendicular magnetic field ( i.e. along north-south direction). But if the material of the rod is paramagnetic or ferromagnetic it will stay in east-west direction having magnetic field in east-west direction.

#### Page No 285:

#### Question 5:

Why is it not possible to make permanent magnets from paramagnetic materials?

#### Answer:

Permanent magnets are made from the material that are easily magnetized and retain the magnetization even reverse magnetizing field is applied (high coercivity). Paramagnetic materials get small magnetization, if they are placed in magnetic field, they lose their magnetization easily if the reverse field is applied. Hence, they are not used to make permanent magnet.

#### Page No 285:

#### Question 6:

Can we have magnetic hysteresis in paramagnetic or diamagnetic substances?

#### Answer:

No, magnetic hysteresis is the lagging of intensity of magnetization (*I*) behind magnetising force (*H*). When diamagnetic and paramagnetic materials are placed in a magnetic field they get weekly magnetised. Also, they lose their magnetization as the magnetic field is removed (low retentively). Therefore, they do not form magnetic hysteresis curve.

#### Page No 285:

#### Question 7:

When a ferromagnetic material goes through a hysteresis loop, its thermal energy is increased. Where does this energy come from?

#### Answer:

When a ferromagnetic material is taken through the cycle of magnetisation, magnet dipoles of the material orient and reorient with time. This molecular motion within the material results in the production of heat, which increses thermal energy of material.

#### Page No 285:

#### Question 8:

What are the advantages of using soft iron as a core, instead of steel, in the coils of galvanometers?

#### Answer:

The material used as a core in the moving coil galvanometer undergoes cycle of magnetization for long period. Therefore, low hysterisis loss is the first requirement for such material. In soft ron core, area under the hysteresis curve is small thus loss of energy is less as compared to steel. Further, it is easily magnetized by the magnetizing field, which increase the magnetic field and hence sensitivity of galvanometer.

#### Page No 285:

#### Question 9:

To keep valuable instruments away from the earth's magnetic field, they are enclosed in iron boxes. Explain.

#### Answer:

As we know that iron have high permeability, therefore it will provide easy path for the magnetic field lines to pass. As a result of this, all the magnetic field lines of earth's magnetic field will prefer to pass through the wall of the box making magnetic field inside the box zero. Hence, it will keep the valuable instruments away from the earth's magnetic field.

#### Page No 286:

#### 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?

#### Answer:

Given:

(a) Intensity of magnetisation, *H* = 1500 A/m

As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, the end effects may be neglected.

The sole effect of the rod in the magnetic field of the solenoid is that a magnetisation will be induced in the rod depending on rod magnetic properties.

There is no effect of the rod on the magnetic intensity at the centre.

(b) Magnetisation of the core, *I *= 0.12 A/m

We know:

$I=\chi H$,

where $\chi $ is the susceptibility of the material of the rod.

$\therefore \chi =\frac{I}{H}=\frac{0.12}{1500}\phantom{\rule{0ex}{0ex}}=0.00008=8\times {10}^{-5}$

(c) The material is paramagnetic.

#### Page No 286:

#### Question 3:

The magnetic field inside a long solenoid of 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 cm^{2} is inserted into it. Find (a) the current in the solenoid (b) the magnetisation *I* of the core and (c) the pole strength developed in the core.

#### Answer:

Given:

Magnetic field strength without iron core, *B*_{1} = 2.5 × 10^{−3} T

Magnetic field after introducing the iron core, *B*_{2} = 2.5 T

Area of cross-section of the iron core, *A* = 4 × 10^{−4} m^{2}

Number of turns per unit length*, n* = 50 turns/cm = 5000 turns/m

(a) Magnetic field produced by a solenoid $\left(B\right)$ is given by,

$B={\mathrm{\mu}}_{0}ni$,

where *i* = electric current in the solenoid

2.5 × 10^{−3} = 4$\mathrm{\pi}$ × 10^{−7}^{ }× 5000 × *i*

$\Rightarrow i=\frac{2.5\times {10}^{-3}}{4\mathrm{\pi}\times {10}^{-7}\times 5000}\phantom{\rule{0ex}{0ex}}=0.398\mathrm{A}=0.4\mathrm{A}\phantom{\rule{0ex}{0ex}}$

$\left(b\right)\mathrm{Magneti}\text{sa}\mathrm{tion}\left(I\right)\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}I\mathit{=}\frac{B}{{\mu}_{\mathit{0}}}\mathit{-}H,\phantom{\rule{0ex}{0ex}}\text{w}\mathrm{here}B\mathrm{is}\mathrm{the}\mathrm{net}\mathrm{magnetic}\mathrm{field}\mathrm{after}\mathrm{introducing}\mathrm{the}\mathrm{core},\mathrm{i}.\mathrm{e}.B=2.5\mathrm{T}.\phantom{\rule{0ex}{0ex}}\mathrm{And}{\mathrm{\mu}}_{0}H\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{magnetising}\mathrm{field},\mathrm{i}.\mathrm{e}.\mathrm{the}\mathrm{diffrence}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{magnetic}\mathrm{fields}\text{'}\mathrm{strengths}.\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{2.5\times {10}^{-3}}{4\mathrm{\pi}\times {10}^{-7}}.\left({B}_{2}-{B}_{1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{2.5\left(1-{\displaystyle \frac{1}{1000}}\right)}{4\mathrm{\pi}\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}\Rightarrow I\approx 2\times {10}^{6}\mathrm{A}/\mathrm{m}$

$\left(\mathrm{c}\right)\mathrm{Intensity}\mathrm{of}\mathrm{magneti}\text{s}\mathrm{ation}\left(I\right)\mathrm{is}\mathrm{given}\mathrm{by},\phantom{\rule{0ex}{0ex}}I\mathit{}\mathit{=}\mathit{}\frac{M}{V}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}I\mathit{=}\mathit{}\frac{m\times 2I}{A\times 2I}\mathit{}\mathit{=}\mathit{}\frac{m}{A}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}m\mathit{}\mathit{=}\mathit{}lA$

* $\Rightarrow $m*= 2 × 10^{6} × 4 × 10^{−4}

* $\Rightarrow $m*= 800 A-m

#### Page No 286:

#### Question 4:

A bar magnet of length 1 cm and cross-sectional area 1.0 cm^{2} 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 magnetisation *I* of the magnet. (c) Find the magnetic field *B* at the centre of the magnet.

#### Answer:

Given:

Distance of the observation point from the centre of the bar magnet, *d* = 15 cm = 0.15 m

Length of the bar magnet, *l* = 1 cm = 0.01 m

Area of cross-section of the bar magnet, *A* = 1.0 cm^{2} = 1 × 10^{−4} m^{2}

Magnetic field strength of the bar magnet,* B *= 1.5 × 10^{−4} T

As the observation point lies at the end-on position, magnetic field $\left(B\right)$ is given by,

$\overrightarrow{B}=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\times \frac{2Md}{({d}^{2}-{l}^{2}{)}^{2}}$

On substituting the respective values, we get:

$1.5\times {10}^{-4}=\frac{{10}^{-7}\times 2\times M\times 0.15}{(0.0225-0.0001{)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1.5\times {10}^{-4}=\frac{3\times {10}^{-8}\times M}{5.01\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{1.5\times {10}^{-4}\times 5.01\times {10}^{-4}}{3\times {10}^{-8}}\phantom{\rule{0ex}{0ex}}=2.5\mathrm{A}$

(b) Intensity of magnetisation (*I*) is given by,

* I* = $\frac{M}{V}$

$=\frac{2.5}{{10}^{-4}\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{6}\mathrm{A}/\mathrm{m}$

(c)

$\mathrm{H}=\frac{\mathrm{M}}{4\mathrm{\pi}l{d}^{2}}$

$=\frac{2.5}{4\times 3.14\times 0.01\times (0.15{)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2.5}{4\times 3.14\times 1\times {10}^{-2}\times 2.25\times {10}^{-2}}$

Net H = H_{N} + H_{S}

= 884.6 = 8.846 × 10^{2}

= 314 T

$\overrightarrow{\mathrm{B}}$ = µ_{0} (H + 1)

= π × 10^{−7} (2.5 × 10^{6} + 2 × 884.6)

= 3.14 T

#### Page No 286:

#### Question 5:

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

#### Answer:

Susceptibility of annealed iron, $\chi $ = 5500

The relation between permeability and susceptibility:

Permeability, *µ* = *µ*_{0}(1 + *x*)

*µ* = 4$\mathrm{\pi}$ × 10^{−7}^{ }(1 + 5500)

*$\Rightarrow $ µ* = 4 × 3.14 × 10^{−7} × 5501

$\Rightarrow $*µ* = 69092.56 × 10^{−7}

$\Rightarrow $*µ* = 6.9 × 10^{−3}

#### Page No 286:

#### 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.

#### Answer:

Here,

Magnetic field strength, *B* = 1.6 T

Magnetising intensity in a material, *H* = 1000 A/m

The relation between magnetic field and magnetising field is given by

$\mu \mathit{=}\frac{B}{H}\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =\frac{1.6}{1000}\phantom{\rule{0ex}{0ex}}\Rightarrow \mu =1.6\times {10}^{-3}$

$\text{R}\mathrm{elative}\mathrm{perm}\text{ea}\mathrm{bility}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{permeability}\mathrm{in}\text{a}\mathrm{medium}\mathrm{to}\mathrm{that}\mathrm{in}\mathrm{vacuum}.\phantom{\rule{0ex}{0ex}}\text{So,}{\mu}_{r}=\frac{\mu}{{\mu}_{0}}=\frac{1.6\times {10}^{-3}}{4\mathrm{\pi}\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{r}=0.127\times {10}^{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{r}=1.3\times {10}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{Relative}\mathrm{permeability}\left({\mu}_{r}\right)\mathrm{is}\mathrm{given}\mathrm{by},\phantom{\rule{0ex}{0ex}}{\mu}_{r}=(1+\chi )\phantom{\rule{0ex}{0ex}}\Rightarrow \chi =1.3\times {10}^{3}-1\phantom{\rule{0ex}{0ex}}\Rightarrow \chi =1300-1\phantom{\rule{0ex}{0ex}}\Rightarrow \chi =1299=1.299\times {10}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \chi \approx 1.3\times {10}^{3}$

#### Page No 286:

#### Question 7:

Electromagnets are made of soft iron because soft iron has

(a) high retentivity and high coercive force

(b) high retentivity and low coercive force

(c) low retentivity and high coercive force

(d) low retentivity and low coercive force

#### Answer:

(d) low retentivity and low coercive force

Electromagnets are made of soft iron because soft iron has

(a) low retentivity - When soft iron is placed inside a solenoid to make an electromagnet and current is passed through the solenoid,magnetism of the solenoid is incresed thousand folds. When the current is switched off, the magnetism is removed instantly because of the low retentivity of soft iron.

(a) Low coercivity - it has low coercivity so that area under the hysteresis cureve for soft iron is very small hysterisis loss in case of soft iron is small.

#### Page No 286:

#### 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 centimetre of the solenoid.

#### Answer:

Here,

Current in the solenoid, *I* = 2 A

Magnetic intensity at the centre of long solenoid, *H = *1500 Am^{−1}

Magnetic field produced by a solenoid$\left(B\right)$ is given by

*B = *µ_{0}*ni** ...*(*1*)

Here,* **n* = number of turns per unit length

* i* = electric current through the solenoid

Also, the relation between magnetic field strength$\left(B\right)$ and magnetic intensity$\left(H\right)$ is given by

*H *= $\frac{B}{{\mathrm{\mu}}_{0}}$ ...(2)

From equations (1) and (2), we get:

*H* = *ni*

⇒ 1500 A/m = *n* × 2

⇒ *n* = 750 turns/meter

⇒ *n* = 7.5 turns/cm

#### Page No 287:

#### 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}?

#### Answer:

Given,

Susceptibility of magnesium at 300 K, ${\chi}_{1}$ = 1.2 × 10^{−5}

Let *T*_{1} be the temperature at which susceptibility of magnesium is 1.2 × 10^{−5} and *T*_{2 }be the temperature at which susceptibility of magnesium is 1.8 × 10^{−5}.

According to Curie's law,

$\chi =\frac{\mathrm{C}}{T},\phantom{\rule{0ex}{0ex}}$

where C is Curie's constant.

$\Rightarrow \frac{{\chi}_{1}}{{\chi}_{2}}=\frac{{T}_{2}}{{T}_{1}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1.2\times {10}^{-5}}{1.8\times {10}^{-5}}=\frac{{T}_{2}}{300}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=\frac{12}{18}\times 300\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{2}=\frac{2}{3}\times 300=200\mathrm{K}$

#### Page No 287:

#### 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 m^{2}). The density of atoms in iron is 8.52 × 10^{28} atoms m^{−3}. (a) Find the maximum magnetisation* I* in a long cylinder of iron (b) Find the maximum magnetic field *B* on the axis inside the cylinder.

#### Answer:

Given:

No of atoms per unit volume, *f* = 8.52 × 10^{28} atoms/m^{3}

Magnetisation per atom, *M* = 2 × 9.27 × 10^{−24} A-m^{2}

(a) Intensity of magnetisation, *I* = $\frac{M}{V}$

*$\Rightarrow $I* = 2 × 9.27 × 10^{−24} × 8.52 × 10^{28}

$\Rightarrow $*I* = 1.58 × 10^{6} A/m.

(b) For maximum magnetisation ,the magnetising field will be equal to the intensity of magnetisation.

So, *I* = *H*

Magnetic field (*B*) will be,

* B* = 4$\mathrm{\pi}$ × 10^{−7} × 1.58 × 10^{6}

* $\Rightarrow $B* ≈ 19.8 × 10^{−1} = 2.0 T.

#### Page No 287:

#### Question 9:

The coercive force for a certain permanent magnet is 4.0 × 10^{4} A m^{−1}. This magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetise it completely. Find the current.

#### Answer:

Given:

Number of turns per unit length, *n* = 40 turns/cm = 4000 turns/m

Magnetising field, *H* = 4 × 10^{4} A/m

Magnetic field inside a solenoid (*B*) is given by,

*B* = *µ*_{0}*n**I**, *

where, *n* = number of turns per unit length.

* I* = current through the solenoid.

$\therefore \frac{B}{{\mu}_{\mathit{0}}}\mathit{}\mathit{=}\mathit{}nI\mathit{}\mathit{=}\mathit{}H$

$\Rightarrow H\mathit{=}\frac{N}{l}I\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}I\mathit{=}\frac{Hl}{N}\mathit{}\mathit{=}\mathit{}\frac{H}{n}\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{4\times {10}^{4}}{4000}=10\mathrm{A}$

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