Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 5 - Magnetism And Matter

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Page No 28:

Question 1:

A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m

(a) is non-zero and points in the z-direction by symmetry.

(b) points along the axis of the tortoid ( m = m

\hat{ϕ}

).
(c) is zero, otherwise there would be a field falling as

\frac{1}{r^{3}}

at large distances outside the toroid.

(d) is pointing radially outwards.

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 $m \propto \frac{1}{r^{3}}$ 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^âˆ˜ with the axis of earth. At Mumbai, declination is nearly zero. Then,
(a) the declination varies between 11.3^âˆ˜ W to 11.3^âˆ˜E.
(b) the least declination is 0^âˆ˜.
(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:

In a permanent magnet at room temperature

(a) magnetic moment of each molecule is zero.

(b) the individual molecules have non-zero magnetic moment which are all perfectly aligned.

(d) domains are all perfectly aligned.

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:

Consider the two idealized systems: (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length L >> R, radius of cross-section. In (i) E is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below:

(a) case (i) contradicts Gauss’s law for electrostatic fields.

(b) case (ii) contradicts Gauss’s law for magnetic fields.
(c) case (i) agrees with

\oint E . d l = 0 .

(d) case (ii) contradicts

\oint H . d l = I_{e n}

Answer:

Gauss law for an electrostatic field is given as $\oint_{s} E . d s = \frac{q}{E_{o}}$
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 $\oint B . d s = 0$
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:

A paramagnetic sample shows a net magnetisation of 8 Am^–1 when placed in an external magnetic field of 0.6 T at a temperature of 4 K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetisation will be

(a) \frac{32}{3} {Am}^{- 1} (b) \frac{2}{3} {Am}^{- 1} (c) 6 {Am}^{- 1} (d) 2.4 {Am}^{- 1} .

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:
$M \propto \frac{B}{T} \Rightarrow \frac{M_{2}}{M_{1}} = \frac{B_{2}}{T_{2}} \times \frac{T_{1}}{B_{1}} = \frac{0.2}{16} \times \frac{4}{0.6} \Rightarrow \frac{M_{2}}{M_{1}} = \frac{1}{12} \Rightarrow M_{2} = 8 \times \frac{1}{12} = \frac{2}{3} {Am}^{- 1}$
Hence, the correct answer is option (b).

Page No 29:

Question 6:

S is the surface of a lump of magnetic material.

(a) Lines of B are necessarily continuous across S.

(b) Some lines of B must be discontinuous across S.

(d) Lines of H cannot all be continuous across S.

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, $H = \frac{B}{μ_{0} μ_{r}}$
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:

The primary origin(s) of magnetism lies in

(a) atomic currents.

(b) Pauli exclusion principle.

(d) intrinsic spin of electron.

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:

A long solenoid has 1000 turns per metre and carries a current of 1 A. It has a soft iron core of

μ_{r}

= 1000. The core is heated beyond the Curie temperature, T_c.

(a) The H field in the solenoid is (nearly) unchanged but the B field decreases drastically.

(b) The H and B fields in the solenoid are nearly unchanged.

(d) The magnetisation in the core diminishes by a factor of about 10⁸.

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.
$\Rightarrow H = n I = 1000 \times 1 = 1000 A / m$
Magnetic induction is the product of permeability and magnetic intensity
$B = μ H = μ_{0} μ_{r} H$
On heating iron core beyond curie temperature value of $μ_{r}$ changes whereas H will remain same and will behave like paramagnetic meterial.
$\Rightarrow (χ_{m}) para = 10^{- 5} and (χ_{m}) ferro = 10^{3} \Rightarrow \frac{{(χ_{m})}_{ferro}}{{(χ_{m})}_{para}} = 10^{8}$
Hence, the correct options are (a) and (d).

Page No 30:

Question 9:

Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to

(a) electrostatic field lines can end on charges and conductors have free charges.

(b) lines of B can also end but conductors cannot end them.

(d) shells of high permeability materials can be used to divert lines of B from the interior region.

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:

Let the magnetic field on earth be modelled by that of a point magnetic dipole at the centre of earth. The angle of dip at a point on the geographical equator

(a) is always zero.

(b) can be zero at specific points.

(d) is bounded.

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 B_v 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 B_v 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:

A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?

Answer:

Magnetic moment due to spinning,
$M = \frac{e h}{4 π m} \Rightarrow M \propto \frac{1}{m} (as, e, h and 4 π are constant)$
So the ratio of magnetic moment of proton and the electron:
$\Rightarrow \frac{M_{p}}{M_{e}} = \frac{m_{e}}{m_{p}} = \frac{m_{e}}{1837 m_{e}} = \frac{1}{1837} (∵ m_{p} = 1837 m_{e}) \Rightarrow \frac{M_{p}}{M_{e}} < < 1 \Rightarrow M_{p} < < M_{e}$
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 I_M.

Answer:

Given: Intensity of magnetisation, M = 10⁶ A/m
Length of a permanent magnet, l = 10 $\times$ 10^–2 = 0.1 m.
$\begin{array}{rcl} Magnetisation current, I_{M} & = & M l \\ = & 10^{6} \times 0.1 \\ = & 10^{5} A \end{array}$
Hence, the value of magnetisation current is 10⁵ A.

Page No 30:

Question 13:

Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of N₂ (~5 × 10^–9) (at STP) and Cu (~10^–5).

Answer:

Given: Magnetic susceptibility of N₂ = 5 $\times$ 10^–9
Magnetic susceptibility of Cu = 10^–5
Density of nitogen, $ρ_{N_{2}} = \frac{28 g}{22.4 L} = \frac{28 g}{22400 cc}$
Density of copper, $ρ_{C u} = 8 g / {cm}^{3}$
$\Rightarrow \frac{ρ_{N_{2}}}{ρ_{C u}} = \frac{28}{22400 \times 8} = 1.6 \times 10^{- 4} Similarly, \frac{χ_{N_{2}}}{χ_{C u}} = \frac{5 \times 10^{- 9}}{10^{- 5}} = 5 \times 10^{- 4}$
Magnetic susceptibility, $χ = \frac{Intensity of magnetisation}{Megnetising field intensity}$
$\Rightarrow χ = \frac{M}{H V} \Rightarrow χ = \frac{M \times ρ}{H m}$
Here, $\frac{M}{m H}$ is constant,
$\Rightarrow χ \propto ρ$
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:

A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. (i) In which direction will it move? (ii) What will be the direction of it’s magnetic moment?

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:

Verify the Gauss’s law for magnetic field of a point dipole of dipole moment m at the origin for the surface which is a sphere of radius R.

Answer:

In accordance with Gauss law of magnetism,
$\oint \vec{B} . \vec{d s} = 0$
Magnetic field induction at P due to dipole,
$\vec{B} = \frac{μ}{4 π} \frac{2 M \cos θ}{r^{3}} \hat{r}$
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.
$\begin{array}{rcl} \vec{d s} & = & r (r \sin θ d θ) \hat{r} \\ = & r^{2} \sin θ d θ \hat{r} \end{array}$
$\Rightarrow ϕ = B . d s = \frac{μ_{0}}{4 π} \frac{2 M \cos θ}{r^{3}} \hat{r} (r^{2} \sin θ d θ \hat{r}) = \frac{μ_{0}}{4 π} \frac{M}{r} \int_{0}^{2 π} 2 \sin θ \cos θ d θ = \frac{μ_{0}}{4 π} \frac{M}{r} \int_{0}^{2 π} \sin 2 θ d θ = \frac{μ_{0}}{4 π} \frac{M}{r} {[\frac{- \cos 2 θ}{2}]}_{0}^{2 π} = \frac{μ_{0}}{4 π} \frac{M}{2 r} [1 - 1] = 0$

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:

Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole p in an electrostatic field E and (ii) magnetic dipole m in a magnetic field B. Write down a set of conditions on E, B, p, m so that the two motions are verified to be identical. (Assume identical initial conditions.)

Answer:

Let the angle between electric dipole p in electrostatic field E be $θ$ .
Torque on electric dipole, $τ = p E \sin θ$
Also, the angle between $\vec{m} and \vec{B} be θ .$
Torque on magnetic dipole moment in magnetic field, $τ' = m B \sin θ$
Two motions will be identieal if $τ = τ'$
$\Rightarrow p E \sin θ = m B \sin θ \Rightarrow p E = m B$
But
$E = c B \Rightarrow p c B = m B \Rightarrow p = \frac{m}{c}$
Hence, these are the set of conditions so that motion are verified to be identical.

Page No 31:

Question 19:

A bar magnet of magnetic moment m and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field B. What would be the similar period T′ for each piece?

Answer:

Magnetic moment = m
Moment of inertia = I $= \frac{M l^{2}}{12}$
Time period due to oscillation, $T = 2 π \sqrt{\frac{I}{m B}}$
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
$m' = \frac{m}{n}$
Magnetic dipole moment for two parts of the magnet,
$m' = \frac{m}{2}$
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
$I' = \frac{(\frac{M}{2}) {(\frac{l}{2})}^{2}}{12} = \frac{1}{8} \times \frac{M l^{2}}{12} = \frac{I}{8}$
$Time period, T' = 2 π \sqrt{\frac{I'}{m' B}} = 2 π \sqrt{\frac{\frac{I}{8}}{\frac{m}{2} \times B}} = \frac{2 π}{2} \sqrt{\frac{I}{m B}} \Rightarrow T' = \frac{T}{2}$
Hence, the time period for each piece is $\frac{T}{2} .$

Page No 31:

Question 20:

Use (i) the Ampere’s law for H and (ii) continuity of lines of B, to conclude that inside a bar magnet, (a) lines of H run from the N pole to S pole, while (b) lines of B must run from the S pole to N pole.

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:
$\int_{P}^{Q} \vec{H} \vec{d l} = \int_{Q}^{P} \frac{\vec{B}}{μ_{0}} . \vec{d l}$
Angle between $\vec{B}$ and $\vec{d l}$ varies continuously.
$\Rightarrow \int_{P}^{Q} \vec{H} . \vec{d l} = \int_{Q}^{p} \frac{\vec{B}}{μ_{0}} \vec{d l} > 0 i . e .,$ positive value of magnetic field lines will move from south pole to north pole inside the magnet.
Since $\underset{P Q P}{∳} \vec{H} \vec{d l} = 0$

$\Rightarrow \underset{P Q P}{∳ \vec{H}} . \vec{d l} = 0$

$Here, \int_{P}^{Q} H . d l > 0 (outside) \int_{P}^{Q} H . d l < 0 (inside)$
It is possible when angle between H and dl is more then 90°, so $\cos θ$ is negative.

Page No 31:

Question 21:

Verify the Ampere’s law for magnetic field of a point dipole of dipole moment m = m $\hat{k}$ . Take C as the closed curve running clockwise along (i) the z-axis from z = a > 0 to z = R; (ii) along the quarter circle of radius R and centre at the origin, in the first quadrant of x-z plane; (iii) along the x-axis from x = R to x = a, and (iv) along the quarter circle of radius a and centre at the origin in the first quadrant of x-z plane.

Answer:

Magnetic field induction, B at a point (0, 0, z) from the magnetic dipole of magnetic moment m,
$|B| = \frac{μ_{0}}{4 π} \frac{2 |m|}{z^{3}} = \frac{μ_{0} m}{2 {πz}^{3}}$
(i) When, the closed curve running clockwise along the z-axis from z = a > 0 to z = R as shown in the diagram:
$\int_{a}^{R} B . d l = \frac{μ_{ο} \times 2 m}{4 π} \int_{a}^{R} \frac{d z}{z^{3}} = \frac{μ_{0} m}{4 π} (\frac{1}{a^{2}} - \frac{1}{R^{2}})$

(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,
$B = \frac{μ_{0}}{4 π} \frac{m \sin θ}{R^{3}}$
Here, $d l = R d θ \Rightarrow d θ = \frac{d l}{R}$
Applying Ampere's law
$\begin{array}{rcl} \int B . d l & = & \int B d l \cos θ = \int_{0}^{\frac{π}{2}} \frac{μ_{0}}{4 π} \frac{m \sin θ}{R^{3}} R d θ \\ = & \int_{0}^{\frac{π}{2}} \frac{μ_{0}}{4 π} \frac{m \sin θ}{R^{2}} d θ \\ = & \frac{μ_{0}}{π} \frac{m}{R} {[- \cos θ]}_{0}^{\frac{π}{2}} \\ = & \frac{μ_{0}}{4 π} \frac{m}{R^{2}} \end{array}$

(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
$B = \frac{μ_{0}}{4 π} \frac{m}{x^{3}}$
Applying Ampere's law, if angle between M and dl is 90°
$\Rightarrow \int_{R}^{a} B . d l = 0$

(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,
$\begin{array}{rcl} \int B . d l & = & \int_{\frac{π}{2}}^{0} \frac{μ_{0}}{4 π} \frac{m \sin θ}{a^{3}} a d θ \\ = & \frac{μ_{0}}{4 π} \frac{m}{a^{2}} \int_{\frac{π}{2}}^{0} \sin θ d θ \\ = & \frac{μ_{0}}{4 π} \frac{m}{a^{2}} {[- \cos θ]}_{\frac{π}{2}}^{0} \\ = & \frac{μ_{0} m}{4 π a^{2}} [- 1 + 0] \\ = & - \frac{μ_{0} m}{4 π a^{2}} \end{array}$
$\begin{array}{rcl} \Rightarrow & \underset{P Q S T}{∳} B . d l = \int_{P}^{Q} B . d l + \int_{Q}^{S} B . d l + \int_{S}^{T} B . d l + \int_{T}^{P} B . d l \\ = & \frac{μ_{0} m}{4 π} (\frac{1}{a^{2}} - \frac{1}{R^{2}}) + \frac{μ_{0}}{4 π} \frac{m}{R^{2}} + 0 + (- \frac{μ_{0}}{4 π} \frac{m}{a^{2}}) \\ = & 0 \end{array}$
$\Rightarrow \underset{P Q S T}{∳} B . d l = 0$

Page No 32:

Question 22:

What are the dimensions of χ, the magnetic susceptibility? Consider an H-atom. Guess an expression for χ, upto a constant by constructing a quantity of dimensions of χ, out of parameters of the atom: e, m, v, R and μ₀. Here, m is the electronic mass, v is electronic velocity, R is Bohr radius. Estimate the number so obtained and compare with the value of

|χ| ~ 10^{- 5}

for many solid materials.

Answer:

Magnetic susceptibility is a measure of how magnetic material responds to an external field.
$χ = \frac{Intensity of magnetisation (H)}{Magnetision force (M)} [M ° L ° T °]$
Since H and M have same units and dimentions. So, $χ$ are dimensionless.
Also $χ$ is related to e, m, v, R and $μ_{0}$
Applying Biot Savart's law
$d B = \frac{μ_{0}}{4 π} \frac{I d l \sin θ}{r^{2}} \Rightarrow μ_{0} = \frac{4 {πr}^{2} dB}{I d l \sin θ} \Rightarrow μ_{0} = \frac{4 {πr}^{2}}{I d l \sin θ} \times \frac{F}{q v \sin θ}$
$\begin{array}{rcl} \Rightarrow & μ_{0} = \frac{[L^{2}] [M L T^{- 2}]}{[Q T^{- 1}] [L] [Q] [L T^{- 1}]} \\ = & [M L Q^{- 2}] \end{array}$
Since $χ$ is dimensionless, it should have no involvement of charge in its dimensional formula. It will be so if $μ_{0} and e$ should have $μ_{0} e^{2}$ where e is dimension of charge.
Let $χ = μ_{0} e^{2} m^{a} v^{b} R^{c}$
$\Rightarrow [M ° L ° T °] = [M L A^{- 2} T^{- 2}] [A^{2} T^{2}] {[M]}^{a} {[L T^{- 1}]}^{b} {[L]}^{e} = [M^{1 + a} L^{1 + b + c} T^{- b} A^{0}]$
On comparing RHS with LHS
$b = 0 1 + a = 0 \Rightarrow a = - 1 1 + b + c = 0 \Rightarrow c = - 1$
$\Rightarrow χ = \frac{μ_{0} e^{2}}{m R}$
Here,
$μ_{0} = 4 π \times 10^{- 7} Tm / A e = 1.6 \times 10^{- 19} m = 9.1 \times 10^{- 31} kg R = 10^{- 10} m$
$\Rightarrow χ = \frac{4 π \times 10^{- 7} \times {(1.6 \times 10^{- 19})}^{2}}{9.1 \times 10^{- 31} \times 10^{- 10}} \approx 10^{- 4}$
Susceptibility of solid material = 10^–5
On comparing, $\frac{χ}{χ_{s o i l d}} = \frac{10^{- 4}}{10^{- 5}} = 10$

Page No 32:

Question 23:

Assume the dipole model for earth’s magnetic field B which is given by B_V= vertical component of magnetic field =

\frac{μ_{0}}{4 π} \frac{2 m \cos θ}{r^{3}}

B_H= Horizontal component of magnetic field =

\frac{μ_{0}}{4 π} \frac{\sin θ m}{r^{3}}

θ = 90° – lattitude as measured from magnetic equator.

Find loci of points for which (i)

| B |

is minimum; (ii) dip angle is zero; and (iii) dip angle is ± 45°.

Answer:

Given: Vertical component of magnetic field,
$B_{v} = \frac{μ_{0}}{4 π} \frac{2 m \cos θ}{r^{2}} . . . . . (1)$
Horizontal component of magnetic field,
$B_{H} = \frac{μ_{0}}{4 π} \frac{\sin θ m}{r^{3}} . . . . . (2)$
On adding and squarin both therms, we get
$\Rightarrow {B_{v}}^{2} + {B_{H}}^{2} = {(\frac{μ_{0}}{4 π})}^{2} \frac{m^{2}}{r^{6}} [4 \cos^{2} θ + \sin^{2} θ]$
$\begin{array}{rcl} Magnetic field, B & = & \sqrt{{B_{v}}^{2} + {B_{H}}^{2}} \\ = & \frac{μ_{0}}{4 π} \frac{m}{r^{3}} {[3 \cos^{2} θ + 1]}^{\frac{1}{2}} \end{array}$
Value of magnetic field is minimum where $\cos θ = \frac{π}{2}$
This means the magnetic equator is the locis.

(b) Angle of dip, $\tan δ = \frac{B_{v}}{B_{H}}$
$\begin{array}{rcl} = & \frac{\frac{μ_{0}}{4 π} \frac{2 m \cos θ}{r^{3}}}{\frac{μ_{0}}{4 π} \frac{\sin θ m}{r^{3}}} = \frac{2 \cos θ}{\sin θ} \\ = & 2 c o t θ \end{array}$

Angle of dip is zero while mean $δ = 0$

$\Rightarrow c o t θ = 0 \Rightarrow θ = \frac{π}{2}$
This means the magnetic equator is the locus.

(c) When angle of dip is 45°
$\Rightarrow \frac{B_{v}}{B_{H}} = \tan 45 ° = 1 \Rightarrow 2 c o t θ = 1 \Rightarrow c o t θ = \frac{1}{2} \Rightarrow \tan θ = 2 \Rightarrow θ = \tan^{- 1} (2)$
this means $θ = \tan^{- 1} (2)$ is the locus.

Page No 32:

Question 24:

Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q.

Answer:

In the above figure, needle is in the north i.e., P is in the plane S.
$\Rightarrow$ 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:

There are two current carrying planar coils made each from identical wires of length L. C₁ is circular (radius R ) and C₂ is square (side a). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform B and carry the same current. Find a in terms of R.

Answer:

Let C₁ be the circular coil of radius R length L and number of turns per unit length be $n_{1} = \frac{L}{2 πR}$

$\begin{array}{rcl} Magnetic moment of C_{1}, m_{1} & = & n_{1} I A_{1} \\ = & \frac{L I π R^{2}}{2 π R} = \frac{L I R}{2} \end{array}$
Moment of inertia of C₁,
$I_{1} = \frac{M R^{2}}{2}$
Time period of C_1,
$T_{1} = 2 π \sqrt{\frac{I_{1}}{m_{1} B}}$
Similarly C₂ be the the square of side a and perimeter L, number of turns per unit length,
$n_{2} = \frac{L}{4 a}$
$\begin{array}{rcl} \Rightarrow & M a g n e t i c m o m e n t o f C_{2} = m_{2} = n_{2} I A_{2} \\ = & \frac{L}{4 a} I a^{2} \\ = & \frac{L I a}{4} \end{array}$
Moment of inertia of C₂,
$I_{2} = \frac{M a^{2}}{12}$
Time period of C_2,
$= 2 π \sqrt{\frac{I^{2}}{m_{2} B}}$
Since, the frequency of both the coils are same, then time period will also be the same:
$\Rightarrow 2 π \sqrt{\frac{I_{1}}{m_{1} B}} = 2 π \sqrt{\frac{I_{2}}{m_{2} B}} \Rightarrow \frac{I_{1}}{m_{1}} = \frac{I_{2}}{m_{2}} \Rightarrow \frac{\frac{M R^{2}}{2}}{\frac{L I R}{2}} = \frac{\frac{M a^{2}}{12}}{\frac{L I a}{4}} \Rightarrow a = 3 R$
Hence, the value of a is 3R.

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