# NCERT solutions for class 12 Physics chapter 5 Magnetism and Matter

## The comparisons of an electric dipole and magnetic dipole:

 Quantity Electrostatics Magnetism Dipole Moment p m The field at the equatorial point of a dipole $\frac{-p}{4\pi\epsilon_0r^3}$ $\frac{\mu m}{4 \pi r^3}$ The field at the axial point of a dipole $\frac{2p}{4\pi\epsilon_0r^3}$ $\frac{-2\mu m}{4\pi r^3}$ The torque due to a dipole in the external field $p*E$ $m*B$ Energy due to a dipole in the external field $-p.E$ $-m.B$

## NCERT solutions for class 12 physics chapter 5 magnetism and matter exercises:

The three independent quantities used to specify the earth’s magnetic field are:

(i) The horizontal component of Earth's magnetic field  ($H_{E}$).

(ii) The magnetic declination (D): It is the angle between the geographic north and the magnetic north at a place.

(iii)The magnetic dip (I): It is the angle between the horizontal plane and the magnetic axis, as observed in the compass

We would expect a greater angle of dip in Britain. The angle of dip increases as the distance from equator increases.

(It is 0 at the equator and 90 degrees at the poles)

If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

The field lines go into the earth at the north magnetic pole and come out from the south magnetic pole and hence Australia being in the southern hemisphere. The magnetic field lines would come out of the ground at Melbourne.

In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

The magnetic field is perpendicular at the poles and the magnetic needle of the compass tends to align with the magnetic field. Therefore the compass will get aligned in the vertical direction if is held vertically at the north pole.

Magnetic field

$B=\frac{\mu_o M}{4\pi R^3}$

substituting the values

$R=6.4\times10^6\ m$

$\mu_0=4\pi \times 10^{-7}$

$M=8 \times 10 ^ { 22} J T ^{-1}$

then

$B=0.3G$

5.2 (a) Answer the following questions

The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

Due to the constant but slow motion of the plates and change in the core, magnetic field due to Earth may change with time too. The time scale is in centuries for appreciable change.

5.2 (b) Answer the following questions

The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

The iron present in the core of the Earth is in the molten form. Hence it loses it ferromagnetism and not regarded by geologists as a source of earth's magnetism.

5.2 (c) Answer the following questions

The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?

The radioactive materials might be the battery to sustain such currents.

5.2 (d) Answer the following questions

The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?

The direction of the earth's magnetic field was recorded in rocks during solidification. By studying them, geologists can tell if the direction of the field had reversed.

5.2 (e) Answer the following questions

The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) due to the presence of ions in the ionosphere. These ions in motion generate magnetic field and hence distort the shape of a magnetic dipole.

5.2  (f) Answer the following questions

Interstellar space has an extremely weak magnetic field of the order of $10 ^ {-12}$. Can such a weak field be of any significant consequence? Explain.

This weak magnetic field can affect the motion of a charged particle in a circular motion. And a small deviation from its path in the vast interstellar space may have huge consequences.

Given,

The angle between axis of bar magnet and external magnetic field, θ = 30°

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, Τ = 4.5 x $\dpi{100} 10^{-2}$J

We know,

Torque experienced by a bar magnet placed in a uniform magnetic field is:

T = m x B = mBsin$\dpi{80} \theta$

$m = \frac{T}{Bsin\theta}$

$\implies m = \frac{4.5\times10^{-2}J}{0.25T\times sin30^{\circ}}$

$\therefore$  m = 0.36 $\dpi{80} JT^{-1}$

Hence, the magnitude of the moment of the Bar magnet is 0.36 $\dpi{80} JT^{-1}$.

Given,

Magnetic moment of magnet, m = 0.32 $\dpi{80} JT^{-1}$

Magnetic field strength, B = 0.15 T

(a) Stable equilibrium: When the magnetic moment is along the magnetic field i.e. $\theta = 0^{\circ}$

(b) Unstable equilibrium: When the magnetic moment is at 180° with the magnetic field i.e. $\theta = 180^{\circ}$

(c) We know that,

U = - m.B = -mBcos$\dpi{80} \theta$

By putting the given values:

U = (-0.32)(0.15)(cos$0^{\circ}$) = -0.048 J

Therefore, Potential energy of the system in stable equilibrium is -0.048 J

Similarly,

U = (-0.32)(0.15)(cos$180^{\circ}$) = 0.048 J

Therefore, Potential energy of the system in unstable equilibrium is 0.048J.

In this case the magnetic field is generated along the axis / length of solenoid so it acts as a magnetic bar.

The magnetic moment is calculated as :-

$M\ =\ NIA$

or                                             $=\ 800\times 3\times 2.5\times 10^{-4}$

or                                              $=\ 0.6\ JT^{-1}$

## 5.6. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Given,

Magnetic field strength, B = 0.25 T

Magnetic moment, m = 0.6 JT−1

The angle between the axis of the solenoid and the direction of the applied field, $\dpi{100} \theta$ =  30°.

We know, the torque acting on the solenoid is:

$\dpi{100} \tau$ = m x B = mBsinθ
= (0.6 $\dpi{80} JT^{-1}$)(0.25 T)(sin 30o

= 0.075 J
= 7.5 x $\dpi{80} 10^{-2}$ J

The magnitude of torque is 7.5 x $\dpi{80} 10^{-2}$ J.

Given.

Magnetic moment, M= 1.5 $\dpi{80} J T^{-1}$

Magnetic field strength, B= 0.22 T

Now,

The initial angle between the axis and the magnetic field, $\theta_1$= 0°

Final angle, $\theta_2$= 90°

We know, The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

$\dpi{100} W = - MB ( cos \theta_2 - cos \theta_1)$

$\dpi{100} \implies W = - (1.5)(0.22) ( cos 90^{\circ} - cos 0^{\circ}) = -0.33(0-1)$
= 0.33 J

5.7 (a)

The amount of work required for the given condition will be:-

$W\ =\ -MB\left [ \cos \Theta _2\ -\ \cos \Theta _1 \right ]$

$W\ =\ -MB\left [ \cos 180^{\circ}\ -\ \cos 0^{\circ} \right ]$

or                                      $=\ 2MB$

or                                     $=\ 2\times1.5\times0.22$

or                                     $=\ 0.66\ J$

For case (i):

$\theta$ = $\theta_2$ = 90°

We know, Torque, $\tau = MBsin\theta$
$= (1.5)(0.22)sin90^{\circ}$
= 0.33 J

For case (ii):

$\theta$ = $\theta_2$ = 180°

We know, Torque,

$\tau = MBsin\theta$
$= (1.5)(0.22)sin180^{\circ}$
= 0

Given,

Number of turns, N = 2000

Area of the cross-section of the solenoid, $A = 1.6$$\dpi{100} 10^{-4} m^2$

Current in the solenoid, I = 4 A

We know, The magnetic moment along the axis of the solenoid is:

m = NIA

= (2000)(4 A)(1.6 x $\dpi{100} 10^{-4} m^2$)

= 1.28 $\dpi{100} Am^2$

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 \times 10 ^{-2} T$  is set up at an angle of 30° with the axis of the solenoid?

Now,

Magnetic field strength, B = $7.5 \times 10 ^{-2} T$

The angle between the magnetic field and the axis of the solenoid, $\theta = 30^{\circ}$

Now, As the Magnetic field is uniform, the Force is zero

Also, we know,

$\tau$ = mxB = mBsinθ

= (1.28 $\dpi{100} JT^{-1}$)($\dpi{100} 7.5 \times 10 ^{-2} T$)(sin $30^{\circ}$)

= 4.8 x $\dpi{100} 10^{-2}$ J

Therefore, Force on the solenoid = 0 and torque on the solenoid = 4.8 x $\dpi{100} 10^{-2}$ J

Given,

Number of turns, N = 16

Radius of the coil, r = 10 cm = 0.1 m

Current in the coil, I = 0.75 A

Magnetic field strength, B = 5.0 x $\dpi{100} 10^{-2}$ T

Frequency of oscillations of the coil, f = 2.0 $\dpi{100} s^{-1}$

Now, Cross-section of the coil, A = $\dpi{100} \pi r^2$ = $\dpi{100} \pi \times (0.1)^2 m^2$

We know, Magnetic moment, m = NIA

= (16)(0.75 A)($\dpi{100} \pi \times (0.1)^2 m^2$

= 0.377 $\dpi{100} JT^{-1}$

We know, frequency of oscillation in a magnetic field is:

$\dpi{100} f = \frac{1}{2\pi}\sqrt{\frac{MB}{I}}$(I = Moment of Inertia of the coil)

$\dpi{100} \implies I = \frac{MB}{4\pi^2f^2}$

$\dpi{100} \implies I = \frac{0.377\times5\times10^{-2}}{4\pi^22^2}$

$I = 1.19\times$ $\dpi{100} 10^{-4}$ $\dpi{100} kgm^2$

The moment of inertia of the coil about its axis of rotation is $1.19$ $\times$ $\dpi{100} 10^{-4}$ $\dpi{100} kgm^2$.

Given,

The horizontal component of earth’s magnetic field, $\dpi{100} B_{H}$ = 0.35 G

Angle made by the needle with the horizontal plane at the place = Angle of dip = $\delta$ = $22 \degree$

We know, $\dpi{100} B_{H}$ = B cos$\delta$, where B is earth's magnetic field

B = $\dpi{100} B_{H}$/cos$\delta$ = 0.35/(cos$22 \degree$) = 0.377 G

The earth’s magnetic field strength at the place is 0.377 G.

Given,

The horizontal component of earth’s magnetic field, BH = 0.16 G

The angle of declination, $\theta$ = $12^{\circ}$

The angle of dip, $\delta$ = $60^{\circ}$

We know, $\dpi{100} B_{H}$ = B cos$\delta$, where B is Earth's magnetic field

B = $\dpi{100} B_{H}$/cos$\delta$ = 0.16/(cos$60 \degree$) = 0.32 G

Earth’s magnetic field is 0.32 G in magnitude lying in the vertical plane, $12^{\circ}$ west of the geographic meridian and $60^{\circ}$ above the horizontal.

Given,

The magnetic moment of the bar magnet, m = 0.48 $\dpi{100} JT^{-1}$

Distance from the centre, d = 10 cm = 0.1 m

We know, The magnetic field at distance d, from the centre of the magnet on the axis is:

$B= \frac{\mu_{0} m}{2\pi r^3}$

$\therefore B= \frac{4\pi\times10^{-7}\times 0.48}{2\pi (0.1)^3}$

$\implies$ $B =$ $0.96 \times 10^{-4} T$

Therefore, the magnetic field on the axis, B = 0.96 G

Note: The magnetic field is along the S−N direction (like a dipole!).

On the equatorial axis,

Distance,d = 10cm = 0.1 m

We know, the magnetic field due to a bar magnet along the equator is:

$B= -\frac{\mu_{0} m}{4\pi d^3}$

$\therefore B=- \frac{4\pi\times10^{-7}\times 0.48}{4\pi (0.1)^3}$

$\implies$ B = $- 0.48 \times 10^{-4} T$

Therefore, the magnetic field on the equatorial axis, B = 0.48 G

The negative sign implies that the magnetic field is along the N−S direction.

Earth’s magnetic field at the given place, B = 0.36 G

The magnetic field at a distance d from the centre of the magnet on its axis is:

$B = \mu_{0}m/2\pi d^{3}$

And the magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

$B' = \mu_{0}m/4\pi d'^3$

= B/2  ( since d' = d, i.e same distance of null points.)

Hence the total magnetic field is B + B' = B + B/2 = (0.36 + 0.18) G = 0.54 G

Therefore, the magnetic field in the direction of earth’s magnetic field is 0.54 G.

Given, d = 14 cm

The magnetic field at a distance d from the centre of the magnet on its axis:

$B = \mu_{0}m/2\pi d^{3}$

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial (perpendicular bisector) line.

The magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

$B = \mu_{0}m/4\pi d'^3$

Equating these two, we get:

$\dpi{100} \\ \frac{1}{2d^3} = \frac{1}{4d\ '^3}\\ \\ \implies \frac{d\ '^3}{d^3} = \frac{1}{2}$

d' = 14 x 0.794 = 11.1cm

The new null points will be at a distance of 11.1 cm on the normal bisector.

Given,

The magnetic moment of the bar magnet, $m = 5.25 \times$ $\dpi{100} 10^{-2}\ JT^{-1}$

The magnitude of earth’s magnetic field at a place, $H = 0.42 G = 0.42 \times$ $\dpi{100} 10^{-4}$   T

The magnetic field at a distance R from the centre of the magnet on the normal bisector is:

$B = \mu_{0}m/4\pi R^3$

When the resultant field is inclined at 45° with earth’s field, B = H

$B = H = 0.42\times$ $10^{-4}$ $T$

$R^3 = \mu_{0}m/4\pi B$  =  $4\pi\times10^{-7}\times5.25\times10^{-2}/4\pi\times0.42\times10^{-4}$

$= 12.5 \times 10^{-5 }m^3$

Therefore,  R = 0.05 m = 5 cm

The magnetic field at a distance R from the centre of the magnet on its axis:

$B = \mu_{0}m/2\pi R^{3}$

When the resultant field is inclined at 45° with earth’s field, B = H

$R^3 = \mu_{0}m/2\pi B$ = $4\pi\times10^{-7}\times5.25\times10^{-2}/2\pi\times0.42\times10^{-4}$= $25 \times 10^{-5}m^3$

Therefore, R = 0.063 m  = 6.3 cm

NCERT solutions for class 12 physics chapter 5 magnetism and matter additional exercises:

At high temperatures, alignment of dipoles gets disturbed due to the random thermal motion of molecules in a paramagnetic sample. But when cooled, this random thermal motion reduces. Hence, a paramagnetic sample displays greater magnetization when cooled.

The magnetism in a diamagnetic substance is due to induced dipole moment. So the random thermal motion of the atoms does not affect it which is dependent on temperature. Hence diamagnetism is almost independent of temperature.

A toroid using bismuth for its core will have slightly greater magnetic field than a toroid with an empty core because bismuth is a diamagnetic substance.

We know that the permeability of ferromagnetic materials is inversely proportional to the applied magnetic field. Therefore it is more for a lower field.

Since the permeability of ferromagnetic material is always greater than one, the magnetic field lines are always nearly normal to the surface of ferromagnetic materials at every point.

Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet for very strong magnetic fields.

According to the graph between B (external magnetic field) and H (magnetic intensity) in ferromagnetic materials, magnetization persists even when the external field is removed. This shows the irreversibility of magnetization in a ferromagnet.

Material that has a greater area of hysteresis loop will dissipate more heat energy. Hence after going through repeated cycles of magnetization, a carbon steel piece dissipates greater heat energy than a soft iron piece, as the carbon steel piece has a greater hysteresis curve area.

Ferromagnets have a record of memory of the magnetisation cycle. Hence it can be used to store memories.

Ceramic, a ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer.

The region can be surrounded by a coil made of soft iron to shield from magnetic fields.

Given,

Current in the cable, I = 2.5 A

Earth’s magnetic field at the location, H = 0.33 G = 0.33 × 10-4T

The angle of dip, $\delta$ = 0

Let the distance of the line of the neutral point from the horizontal cable = r m.

The magnetic field at the neutral point due to current carrying cable is:

$H_{n} = \mu_{0}I/2\pi r$ ,

We know, Horizontal component of earth’s magnetic field, $H_{E}$ = $Hcos$$\delta$

Also, at neutral points, $H_{E}= H_n$

⇒$Hcos$$\delta$$\mu_{0}I/2\pi r$

$\Rightarrow 0.33 \times 10^{-4} T � cos0^o =\frac{4\pi \times 10^{-7\times2.5}}{2\pi r}$

$\Rightarrow r = 1.515 cm$

Required distance is 1.515 cm.

Number of long straight horizontal wires = 4

The current carried by each wire = 1A

earth’s magnetic field at the place = 0.39 G

the angle of dip = 350

magnetic field due to infinite current-carrying straight wire

$B'=\frac{\mu_0I}{2\pi r}$

r=4cm =0.04 m

$B'=\frac{4\pi\times10^{-7}\times1}{2\pi \times4\times10^{-2}}$

magnetic field due to such 4 wires

$B=4\times \frac{4\pi\times10^{-7}\times1}{2\pi \times4\times10^{-2}}=2\times10^{-5}T$

The horizontal component of the earth's magnetic field

$H=0.39\times10^{-4}cos35=0.319\times10^{-4}T=3.19\times10^{-5}T$

the horizontal component of the earth's magnetic field

$V=0.39\times10^{-4}sin35=0.22\times10^{-4}T=2.2\times10^{-5}T$

At the point below the cable

$H'=H-B=3.19\times10^{-5}-2\times10^{-5}=1.19\times10^{-5}T$

The resulting field is

$\sqrt{H'^2+V^2}=\sqrt{(1.19\times10^{-5})^2+(2.2\times10^{-5})^2}=2.5\times10^{-5}T=0.25G$

Given,

Number of turns in the coil, n = 30

Radius of coil, r = 12cm = 0.12m

Current in the coil, I = 0.35A

The angle of dip, $\delta$ = 45o

We know, Magnetic fields due to current carrying coils, B = $\mu_{0}nI/2r$

$B =$ $4\pi \times10^{-7}\times30\times0.35/2\times0.12$

$= 5.49 \times$ $10^{-5}$$T$

Now, Horizontal component of the earth’s magnetic field = Bsin$\delta$

$= 5.49 \times$ $10^{-5}$$T$$� sin45^o$

$=$$3.88\times$$10^{-5}$$T$ (Hint: Take sin45as 0.7)

$=0.388 G$

When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of $90 \degree$ in the anticlockwise sense looking from above, then the needle will reverse its direction. The new direction will be from east to west.

Given,

The magnitude of the first magnetic field, B1 = 1.2 × 10–2 T

The angle between the magnetic field directions, $\dpi{100} \theta$ = 60°

The angle between the dipole and the magnetic field $\dpi{100} B_{1}$ is $\dpi{100} \theta_{1}$ = 15°

Let Bbe the magnitude of the second magnetic field and M be the magnetic dipole moment

Therefore, the angle between the dipole and the magnetic field B2 is $\dpi{100} \theta_{2}$ = $\dpi{100} \theta - \theta_{1}$= 45°

Now, at rotational equilibrium,

The torque due to field B1 = Torque due to field B2

$MB_{1}sin\theta_{1}= MB_{2}sin\theta_{2}$

$B_{2} = \frac{MB_{1}sin\theta_{1}}{Msin\theta_{2}} = \frac{1.2\times10^{-2}\times sin15\degree}{sin45\degree}$

$= 4.39 \times$ $\dpi{100} 10^{-3}$$T$

Hence the magnitude of the second magnetic field $= 4.39 \times$$\dpi{100} 10^{-3}$$T$

The energy of electron beam   =   18 eV

We can write:-

$E\ =\ \frac{1}{2} mv^2$

so                                                     $v\ =\ \sqrt{\frac{2E}{M}}$

We are given horizontal magentic field :          B  = 0.40 G

Also,                                                $Bev\ =\ \frac{mv^2}{r}$

We obtain,                                         $r\ =\ \frac{1}{Be}\sqrt{2EM}$

or                                                       $r\ =\ 11.3\ m$

Using geometry, we can write :-

$\sin \Theta =\ \frac{x}{r} =\ \frac{0.3}{11.3}$

and                                                   $y\ =\ r-r \cos\Theta$

or                                                           $=\ r(1- \sqrt{1- \sin^2\Theta })$

or                                                     $y\ \approx \ 4mm$

Given,

Magnetic field, $\dpi{100} B_{1}$ = 0.64 T

Temperature, $\dpi{100} \theta_{1}$ = 4.2K

And, saturation = 15%

Hence, Effective dipole moment, $\dpi{80} M_{1}$ = 15% of Total dipole moment

$\dpi{80} M_{1}$ = 0.15 x (no. of atomic dipole × individual dipole moment)

$\dpi{100} M_{1}$ = $\dpi{100} 0.15 \times 2 \times 10^{24} \times 1.5 \times 10^{-23}$= 4.5 $\dpi{80} JT^{-1}$

Now,  Magnetic field, $\dpi{100} B_{2}$ = 0.98 T and Temperature, $\dpi{100} \theta_{2}$ = 2.8 K

Let $\dpi{80} M_{2}$ be the new dipole moment.

We know that according to Curie’s Law, $\dpi{80} M\propto \frac{B}{\theta}$

∴ The ratio of magnetic dipole moments

$\\\frac{M_{2}}{M_{1}}= \frac{B_{2}\times \theta_{1}}{B_{1}\times \theta_{2}} \\ \implies M_{2}= \frac{B_{2}\times \theta_{1}}{B_{1}\times \theta_{1}}\times M_{1} \\\implies M_{2}= \frac{0.98T\times4.2K}{0.64T\times2.8K}\times 4.5 JT^{-1}$ $= 10.336$$JT^{-1}$

Therefore, the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 $\dpi{80} JT^{-1}$

Given,

Radius of ring, r = 15cm = 0.15m

Number of turns in the ring, n = 3500

Relative permeability of the ferromagnetic core, $\dpi{100} \mu_{r}$ = 800

Current in the Rowland ring, I = 1.2A

We know,

Magnetic Field due to a circular coil, B  = $\dpi{100} \frac{\mu_{r} \mu_{0}nI}{2\pi r}$

∴ B = $\dpi{100} \frac{4\pi\times10^{-7}\times800\times3500\times1.2}{2\pi \times0.15}$ = 4.48T

Therefore, the magnetic field B in the core for a magnetising current is 4.48 T

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

We know,

$\mu_{l} = -(\frac{e}{2m})l$

$\therefore \mu_{l} = -(\frac{e}{2m})l$ is in expected from classical physics.

Now, Magnetic moment associated with the orbital motion of the electron is:

$\mu_{l}$ = Current x Area covered by orbit = I x A

$(\frac{e}{T})\pi r^2$

And, l = angular momentum = mvr

= $m(\frac{2\pi r}{T}) r$

(m is the mass of the electron having charge (-e), r is the radius of the orbit of by the electron around the nucleus and T is the time period.)

Dividing these two equations:

$\frac{\mu_{l}}{l} = -\frac{e}{T}\pi r^2\times \frac{T}{m\times2\pi r^2} = -\frac{e}{2m}$

$\mu_{l} = (-\frac{e}{2m})l$ , which is the same result predicted by quantum theory.

The negative sign implies that $\mu_{l}$ and l are anti-parallel.

## NCERT Solutions for Class 12 Physics- Chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

NCERT Solutions for Class 12:

## Importance of NCERT solutions for class 12 physics chapter 5 magnetism and matter in exams:

In 2019 CBSE board physics paper, 3 marks questions are asked from the chapter magnetism and matter. Solutions of NCERT class 12 physics chapter 5 magnetism and matter is important for competitive exams like NEET and JEE Main also. If you combine chapters 3 and 4, maybe around 2-4 questions can come in NEET exam and 2-3 questions can come in JEE mains exam. From the CBSE NCERT solutions for class 12 physics chapter 5 magnetism and matter, you can get these marks in your pocket easily.

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