Physics Ncert_exemplar 2019 Solutions for Class 12 Science Physics Chapter 5 Magnetism And Matter are provided here with simple step-by-step explanations. These solutions for Magnetism And Matter are extremely popular among Class 12 Science students for Physics Magnetism And Matter Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert_exemplar 2019 Book of Class 12 Science Physics Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert_exemplar 2019 Solutions. All Physics Ncert_exemplar 2019 Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.
Page No 28:
Question 1:
(c) is zero, otherwise there would be a field falling as at large distances outside the toroid.
Answer:
Toroid is considered as a ring shaped closed solenoid which is like an endless cylindrical solenoid magnetic field is only confined inside the toroid in the form of concentric magnetic lines of force. However, outside the toroid magnetic field is zero because net current enclosed in these spaces is zero. So, magnetic moment of toroid is zero. If we take r as a long distance outside the toroid but this is not possible.
Hence, the correct answer is option (C).
Page No 28:
Question 2:
The magnetic field of Earth can be modelled by that of a point dipole placed at the centre on the Earth. The dipole axis makes an angle of 11.3∘ with the axis of earth. At Mumbai, declination is nearly zero. Then,
(a) the declination varies between 11.3∘ W to 11.3∘E.
(b) the least declination is 0∘.
(c) the plane defined by dipole axis and Earth passes through Greenwich.
(d) declination averaged over Earth must be always negative.
Answer:
The magnetic field lines of Earth is similar to the hypothetical magnetic dipole which is at the center of the Earth. The axis of dipole does not coincide with the axis of rotation of the Earth and is tilled at some angle.
Here, angle of declination is 11.3° so, there are two possibilities:
i.e., declination varies between 11.3°W to 11.3°E.
Hence, the correct answer is option (a).
Page No 29:
Question 3:
Answer:
At room temperature, a permanent magnet behaves like a ferromagnetic substance for a long period of time. The individual atoms possess a dipole movement like in paramagnetic material. They interact with one another in such a way that they spontaneously align themselves in common direction over a macroscopic volume called domain. Theis means domains are perfectly aligned.
Hence, the correct answer is option (d).
Page No 29:
Question 4:
(c) case (i) agrees with
(d) case (ii) contradicts
Answer:
Gauss law for an electrostatic field is given as
Electric field lines do not form a continuous path while magnetic field lines form a closed path.
Gauss law for magnetic field is given as
Thus it contradicts the magnetic field because there is a magnetic field inside the solenoid and no field outside. However, the magnetic field lines form a closed path.
Hence, the correct answer is option B.
Page No 29:
Question 5:
Answer:
Given: Initial intensity of magnetisation,
M1 = 8 Am-1
Initial magnetic field, B1 = 0.6 T
Initial temperature, T1 = 4 K
Final magnetic field, B2 = 0.2 T
Final temperature, T2 = 16 K
Applying Curie's law, the intensity of magnetisation:
Hence, the correct answer is option (b).
Page No 29:
Question 6:
S is the surface of a lump of magnetic material.
Answer:
For any magnet, magnetic field lines form continuous closed loops, so lines of B are necessarily continuous across S.
Magnetic intensity inside the lump,
Since it varies from inside and outside the lump, the lines of H can not be all continuous across S.
Hence, the correct options are (a) and (d).
Page No 30:
Question 7:
Answer:
In an atom, electrons revolve and spin around the nucleus which in turn produces current and due to magnetic effect of current, magnetism is produced.
Hence, the correct options are (a) and (d).
Page No 30:
Question 8:
Answer:
Number of turns per metre in a solenoid, n = 1000
Current through a solenoid, I = 1 A
Magnetic intensity is the product of number of turns per unit length and current.
Magnetic induction is the product of permeability and magnetic intensity
On heating iron core beyond curie temperature value of changes whereas H will remain same and will behave like paramagnetic meterial.
Hence, the correct options are (a) and (d).
Page No 30:
Question 9:
Answer:
The phenomenon which blocks the effect of electric field is known electrostatic shielding. The effect of an external field can be blocked by the conducting shell on its internal content or the effect of internal field outside environment. However, magnetostatic shielding is done by using high permeability magnetic material to prevent static magnetic field not to reach with in the enclosure.
Hence, the correct options are (a), (c) and (d).
Page No 30:
Question 10:
Answer:
The magnetic equator crosses the geographical equator and at those points angle of dip is zero. The geographical equator lies in magnetic north of the magnetic equator and will have Bv in vertically upward, therefore the angle of dip will be negative. The geographical equator lying in magnetic south of the magnetic equator and will have Bv in vertically downward.
Thus, the angle of dip will take maximum to minimum value. So, the angle of dip is also bounded.
Hence, the correct options are (b), (c) and (d).
Page No 30:
Question 11:
Answer:
Magnetic moment due to spinning,
So the ratio of magnetic moment of proton and the electron:
Hence, the magnetic moment of the proton is neglected as compared to that of an electron.
Page No 30:
Question 12:
A permanent magnet in the shape of a thin cylinder of length 10 cm has M = 106 A/m. Calculate the magnetisation current IM.
Answer:
Given: Intensity of magnetisation, M = 106 A/m
Length of a permanent magnet, l = 10 10–2 = 0.1 m.
Hence, the value of magnetisation current is 105 A.
Page No 30:
Question 13:
Answer:
Given: Magnetic susceptibility of N2 = 5 10–9
Magnetic susceptibility of Cu = 10–5
Density of nitogen,
Density of copper,
Magnetic susceptibility,
Here, is constant,
Hence, major difference is accounted for by density.
Page No 31:
Question 14:
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism.
Answer:
The temperature dependence of susceptibility for diamagnetism is not much affected this is because in diamagnetism direction of external magnetic field and magnetism due to orbital motion of electrons are opposite. So, net magnetism becomes zero.
However for paramagnetic and ferromagnetic material if the temperature is raised, the alignment of atomic magnetism is disturbed this is because the direction of magnetism due to the orbital motion of electrons and external applied field is in the same direction. So. net magnetism increases and gets affected by temperature resulting in a decrease in susceptibility.
Page No 31:
Question 15:
Answer:
When a ball of superconducting material is dipped in liquid nitrogen it will remain as diamagnetic material.
On applying an external magnetic field on superconducting material dipped in liquid nitrogen:
(i) it will be repelled by an external magnetic field and hence will move away from the magnet.
(ii) the direction of motion will be opposite to the direction of the magnet or magnetic field.
Page No 31:
Question 16:
Answer:
In accordance with Gauss law of magnetism,
Magnetic field induction at P due to dipole,
Consider a spherical surface of radius r with center O lying in y-z plane.
Let ds be the elementary area of the surface at P.
Hence proved.
Page No 31:
Question 17:
Three identical bar magnets are rivetted together at centre in the same plane as shown in the given figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the given figure. Determine the poles of the remaining two.
Answer:
The system will be in stable equilibrium when the north pole of the magnet one is equally attached by the south pole of the magnet. Also, the net force and torque on the system is zero. So, this could only be possible if poles of remaining two magnets are as shown below:
Page No 31:
Question 18:
Answer:
Let the angle between electric dipole p in electrostatic field E be .
Torque on electric dipole,
Also, the angle between
Torque on magnetic dipole moment in magnetic field,
Two motions will be identieal if
But
Hence, these are the set of conditions so that motion are verified to be identical.
Page No 31:
Question 19:
Answer:
Magnetic moment = m
Moment of inertia = I
Time period due to oscillation,
Here, m is the magnetic moment of the magnet.
B is the uniform magnetic field.
When a magnet is cut into n pieces in equal proportion then the magnetic moment m' of all equal parts is
Magnetic dipole moment for two parts of the magnet,
When magnet is cut into two equal parts, then the moment of inertia of each part of magnet about an axis perpendicular to length passing though its center is
Hence, the time period for each piece is
Page No 31:
Question 20:
Answer:
Let us consider a bar magnet and the magnetic field B forms a closed loop. Consider a loop C inside and outside magnet NS on side PQ of the magnet.
Applying Ampere's law:
Angle between and varies continuously.
positive value of magnetic field lines will move from south pole to north pole inside the magnet.
Since
It is possible when angle between H and dl is more then 90°, so is negative.
Page No 31:
Question 21:
Answer:
Magnetic field induction, B at a point (0, 0, z) from the magnetic dipole of magnetic moment m,
(i) When, the closed curve running clockwise along the z-axis from z = a > 0 to z = R as shown in the diagram:
(ii) When, the closed curve running clockwise along the quarter circle of radius R and centre at the origin, in the first quadrant of x-z plane:
Magnetic field due to element dl,
Here,
Applying Ampere's law
(iii) When, the closed curve running clockwise along the x-axis from x = R to x = a:
Magnetic field induction at distance x from the dipole is
Applying Ampere's law, if angle between M and dl is 90°
(iv) When, the closed curve running clockwise along the quarter circle of radius a and centre at the origin in the first quadrant of x-z plane:
Applying Ampere's law,
Page No 32:
Question 22:
Answer:
Magnetic susceptibility is a measure of how magnetic material responds to an external field.
Since H and M have same units and dimentions. So, are dimensionless.
Also is related to e, m, v, R and
Applying Biot Savart's law
Since is dimensionless, it should have no involvement of charge in its dimensional formula. It will be so if should have where e is dimension of charge.
Let
On comparing RHS with LHS
Here,
Susceptibility of solid material = 10–5
On comparing,
Page No 32:
Question 23:
Answer:
Given: Vertical component of magnetic field,
Horizontal component of magnetic field,
On adding and squarin both therms, we get
Value of magnetic field is minimum where
This means the magnetic equator is the locis.
(b) Angle of dip,
Angle of dip is zero while mean
This means the magnetic equator is the locus.
(c) When angle of dip is 45°
this means is the locus.
Page No 32:
Question 24:
Answer:
In the above figure, needle is in the north i.e., P is in the plane S.
Declination = 0.
Since P is also on magnetic equator which means angle of dip will also be equal to zero.
However, in the second case, Q is also on the magnetic equator which means angle of dip is zero. But since Earth is tilted on its axis by 11.3° thus declination at Q will be 11.3° itself.
Page No 32:
Question 25:
Answer:
Let C1 be the circular coil of radius R length L and number of turns per unit length be
Moment of inertia of C1,
Time period of C1,
Similarly C2 be the the square of side a and perimeter L, number of turns per unit length,
Moment of inertia of C2,
Time period of C2,
Since, the frequency of both the coils are same, then time period will also be the same:
Hence, the value of a is 3R.
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