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:

When a dielectric is placed in an electric field, it gets polarised. The electric field in a polarised material is less than the applied field. When a paramagnetic substance is kept in a magnetic field, the field in the substance is more than the applied field. Explain the reason of this opposite behaviour.

#### Answer:

This opposite behaviour is due to the opposite behaviour of magnetic dipole as compared to electric dipole. When the paramagnetic substance is kept in magnetic field, the direction of magnetic field at the centre of magnetic dipole of substance is along the direction of magnetic moment which is opposite to the case of dipole in electric field. Also, when paramagnetic substance is kept in the magnetic field then its magnetic dipole aligns in the direction of field. Thus, magnetic field due to the magnetic dipole adds up to the applied magnetic field.

Hence, an extra magnetic field produced in the direction of applied field.

#### 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 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 286:

#### Question 2:

A paramagnetic material is kept in a magnetic field. The field is increased till the magnetisation becomes constant. If the temperature is now decreased, the magnetisation

(a) will increase

(b) will decrease

(c) will remain constant

(d) may increase or decrease

#### Answer:

(c) will remain constant

Magnetisation becomes constant i.e all the magnetic moments have got aligned in the direction of the applied field. So now, if the temperature is decreased, thermal vibration of the paramagnetic material will reduce, But as all the magnetic moments are already aligned in the direction of the field so no further alignment can take place due to reduced thermal motion. Thus, there will be no negative effect of decreasing the temperature on the magnetisation. Hence correct option is (c).

#### Page No 286:

#### Question 3:

A ferromagnetic material is placed in an external magnetic field. The magnetic domains

(a) increase in size

(b) decrease in size

(c) may increase or decrease in size

(d) have no relation with the field

#### Answer:

(c) may increase or decrease in size

Atoms of ferromagnetic material in unmagnetised state form domains inside the ferromagnetic material. These domains have large magnetic moments due to the magnetic moment of atoms. In the absence of magnetic field, these domains have magnetic moment in different directions. But when the magnetic field is applied, domains aligned in the direction of the field grow in size and those aligned in the direction opposite to the field reduce in size. Hence, option (c) is correct.

#### Page No 286:

#### Question 4:

A long, straight wire carries a current *i*. The magnetising field intensity *H* is measured at a point *P* close to the wire. A long, cylindrical iron rod is brought close to the wire, so that the point *P *is at the centre of the rod. The value of *H* at *P* will

(a) increase many times

(b) decrease many times

(c) remain almost constant

(d) become zero

#### Answer:

(c) remain almost constant

From the Biot-Savart law, magnetic field $\left(B\right)$ at a point P close to the wire carrying current *i* is given by,

$\overrightarrow{B}=\frac{{\mu}_{0}}{4\mathrm{\pi}}\frac{i\overrightarrow{dl}\times \overrightarrow{r}}{{r}^{3}}$

Magnetising field intensity (*H*) will be,

$H=\frac{B}{{\mu}_{0}}=\frac{1}{4\mathrm{\pi}}\frac{i\overrightarrow{dl}\times \overrightarrow{r}}{{r}^{3}}$

Now, as the cylindrical rod is brought close the wire such that centre of the rod is at P, then distance of point P from the wire(*r*) will remain same. Hence, magnetic field intensity will remain almost constant. Also even when the rod is carrying any current then *B* will be zero at the centre of the rod so the value of Magnetising field intensity will remain the same at point P.

#### Page No 286:

#### Question 5:

The magnetic susceptibility is negative for

(a) paramagnetic materials only

(b) diamagnetic materials only

(c) ferromagnetic materials only

(d) paramagnetic and ferromagnetic materials

#### Answer:

(b) diamagnetic materials only

Magnetic susceptibility is defined as the ratio of the intensity of magnetisation induced in the material to magnetising foorce applied on it.

Magnetic susceptibility is negative for diamagnetic materials. Magnetic susceptibility is positive for paramagnetic and ferromagnetic materials.

#### Page No 286:

#### Question 6:

The desirable properties for making permanent magnets are

(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:

(a) high retentivity and high coercive force

Permanent magnets should have high retentivity so that the magnet is strong and high coercive force, so that magnetisation is not erased by stary magnetic fields, temperature change or due to rough handling etc.

#### 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:

Pick the correct options.

(a) All electrons have magnetic moment.

(b) All protons have magnetic moment.

(c) All nuclei have magnetic moment.

(d) All atoms have magnetic moment.

#### Answer:

(a) All electrons have magnetic moment.

(b) All protons have magnetic moment.

Electrons of an atom moves in circular path around the nucleus and constitute electric current. Since a current loop has magnetic moment, therefore electron also has magnetic moment due to this orbit motion. It also has magnetic moment due to the spinning about its own axis.

Orbital motion of electron around the nucleus give rise to magnetic field around the proton (nucleus). This create a torque and thus magnetic dipole moment on proton.

All nuclei also have magnetic moment but it is several thousand times smaller than the magnetic moment of the electron so it can be ignored in comparison to the magnetic moment of an eletron.

Generally, an atom has no magnetic moment. Because the magnetic moments of electrons of an atom have a tendency to cancel in pairs.

#### Page No 286:

#### Question 2:

The permanent magnetic moment of the atoms of a material is not zero. The material

(a) must be paramagnetic

(b) must be diamagnetic

(c) must be ferromagnetic

(d) may be paramagnetic

#### Answer:

(d) may be paramagnetic

Diamagnetic material have zero magnetic moment on their own. Hence, option (a) is incorrect.

Paramagnetic and ferromagnetic materials have non zero permanent magnetic moment. But we are not sure about the material, whether it is Paramagnetic or ferromagnetic. If the material has large value of permanent magnetic moment, then it will be ferromagnetic and if it has small value of permanent magnetic moment, then it will be Paramagnetic.

Hence, option (d) is correct.

#### Page No 286:

#### Question 3:

The permanent magnetic moment of the atoms of a material is zero. The material

(a) must be paramagnetic

(b) must be diamagnetic

(c) must be ferromagnetic

(d) may be paramagnetic

#### Answer:

(b) must be diamagnetic

Paramagnetic and ferromagnetic materials have non zero permanent magnetic moment on their own. Only atoms of diamagnetic materials have zero permanent magnetic moment. Hence, (b) is correct.

#### Page No 286:

#### Question 4:

Which of the following pairs has quantities of the same dimensions?

(a) Magnetic field *B* and magnetising field intensity *H*

(b) Magnetic field *B* and intensity of magnetisation *I*

(c) Magnetising field intensity *H* and intensity of magnetisation *I*

(d) Longitudinal strain and magnetic susceptibility

#### Answer:

(c) Magnetising field intensity *H* and intensity of magnetisation *I*

(d) Longitudinal strain and magnetic susceptibility

Dimension of Magnetic field *B *is given by,* *

*F=Bqv*

$\Rightarrow $$B=\frac{F}{qv}$

$\Rightarrow $$\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{AT}\right]\left[{\mathrm{LT}}^{-1}\right]}=\left[{\mathrm{MT}}^{-2}{\mathrm{A}}^{-1}\right]$

Dimension of magnetising field intensity *H *is given by*,*

*H*=$\left[\frac{Idl}{{r}^{2}}\right]$

$\Rightarrow $$\frac{\left[\mathrm{AL}\right]}{\left[{\mathrm{L}}^{2}\right]}=\left[{\mathrm{AL}}^{-1}\right]$

Dimension of intensity of magnetisation *I *is also $\left[{\mathrm{AL}}^{-1}\right]$ as both Magnetising field intensity *H* and intensity of magnetisation *I *are measured in* *Am^{$-1$}.

Hence, option (a) and option (b) is incorrect. Option (c) is correct.

As Longitudinal strain and magnetic susceptibility both are dimensionless quantity. Hence, option (d) is correct.

#### Page No 286:

#### Question 5:

When a ferromagnetic material goes through a hysteresis loop, the magnetic susceptibility

(a) has a fixed value

(b) may be zero

(c) may be infinity

(d) may be negative

#### Answer:

(b) may be zero

(c) may be infinity

(d) may be negative

We know,

$\mathrm{Magnetic}\mathrm{susceptibility}=\frac{I}{H}$

At point B, the value of *H* is zero but *I* is non zero. Magnetic susceptibility is infinity here. At point C, value of magnetic susceptibility will be negative. Here Magnetic field is applied in opposite direction to reduce the intensity of magnetization to zero. The applied field to reduce the residual magnetization to zero is called coercivity. At point D, *I *is zero but *H* is not zero so susceptibility of the material will be zero.

#### Page No 286:

#### Question 6:

Mark out the correct options.

(a) Diamagnetism occurs in all materials.

(b) Diamagnetism results from the partial alignment of permanent magnetic moment.

(c) The magnetising field intensity, *H,* is always zero in free space.

(d) The magnetic field of induced magnetic moment is opposite the applied field.

#### Answer:

(a) Diamagnetism occurs in all materials.

(d) The magnetic field of induced magnetic moment is opposite the applied field.

When a material is placed in magnetic field, dipole moment are induced in the atoms by the applied magnetic field. Since the direction of magnetic field due to induced dipole moment is opposite to the applied magnetic field. Therefore, 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. Hence, option (a) and (d) are correct.

Diamagnetic material do not have permanent magnetic moment on their own. When they are placed in magnetic field, dipole moments are induced by the applied magnetic field. Thus, there is no net alignment of permanent magnetic moment so these mterials do not have any permanenet magnetic momentof their own. Hence, option (b) is incorrect.

Magnetic field intensity is not zero in free space. Hence, option (c) is incorrect.

#### 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 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 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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