Voltage sensitivity is \(V_{S}=\frac{I_{S}}{R_{g}}\) 
When the resistance is doubled, then new resistance is \(R_{g}^{\prime}=2 R_{g}\) 
Increase in current sensitivity is \(I_{S}^{\prime}=\left(1+\frac{50}{100}\right) I_{S}=\frac{3}{2} I_{S}\) 
The new voltage sensitivity is \(V_{S}^{\prime}=\frac{\frac{3}{2} I_{S}}{2 R_{g}}=\frac{3}{4} V_{S}\) 
Hence the voltage sensitivity decreases. The percentage decrease in voltage sensitivity is \(\frac{V_{S}-V_{S}^{\prime}}{V_{S}} \times 100 \%=25 \%\).

In the presence of external magnetic field, two processes take place.
(i) The domains having magnetic moments parallel to the field grow bigger in size.
(ii) The other domains (not parallel to field) are rotated, so that they are aligned with the field.

The properties of diamagnetic materials are:
(i) Magnetic susceptibility is negative and it is temperature independent.
(ii) Relative permeability is slightly less than unity.
(iii) The magnetic field lines are repelled or expelled from diamagnetic materials when placed in a magnetic field.
(iv) Example: Copper, Bismuth.
The properties of paramagnetic materials are:
(i) Magnetic susceptibility is positive and small and it is inversely proportional to the temperature.
(ii) Relative permeability is greater than unity.
(iii) The magnetic field lines are attracted into the paramagnetic materials when placed in a magnetic field.
(iv) Example: Chromium, Aluminium.
The properties of ferromagnetic materials are:
(i) Magnetic susceptibility is positive and large and it is inversely proportional to the temperature.
(ii) Relative permeability is large.
(iii) The magnetic field lines are strongly attracted into the ferromagnetic materials when placed in a magnetic field.
(iv) Example: Nickel, Iron

If a charged particle moves in a region of uniform magnetic field such that its velocity is not perpendicular to the magnetic field, then the velocity of the particle is split up into two components: one component is parallel to the field while the other component perpendicular to the field. The component of velocity parallel to field remains unchanged and the component perpendicular to the field keeps changing due to Lorentz force. Hence, the path of the particle is not a circle, it is a helix around the field lines.

It is an arrangement of eletric field (E) and magnetic field (B) perpendicular to each other. When charged particles enter that region, particles with a certain velocity can pass through that region.
v = E/B
The speed is independent of charge and mass.

An ammeter is connected in series with the circuit because the purpose of the ammeter is to measure the current through the circuit. Since the ammeter is a low impedance device connecting it in parallel with the circuit would cause a short circuit, damaging the ammeter and the circuit.

(i) Stretch out forefinger, the middle finger and the thumb of the left hand such that they are in three mutually perpendicular directions.
(ii) If the forefinger points in the direction of magnetic field, the middle finger in the direction of the electric current, then thumb will point in the direction of the force experienced by the conductor.

Resonance condition happens, when the frequency f at which the positive ion circulates in the magtetic field must be equal to the constant frequency of the electrical oscillator f_osc.
 \(\mathrm{f}_{\mathrm{osc}}=\frac{\mathrm{q} \mathrm{B}}{2 \pi \mathrm{m}}\)

(i) Magnetic declination is the angle between magnetic meridian at a point and geographical meridian.
(ii) Magnetic inclination or dip at a point is defined as the angle subtended by the Earth's total magnetic field \(\vec{B}\) with the horizontal direction in the magnetic meridian.

One ampere is defined as that constant current when it is passed through each of the two infinitely long parallel straight conductors kept side by side parallely at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 x 10^-7 newton per meter length of conductor.

Magnetic field B = 0.2 T
Current flowing through the coil I = 3A
Cross sectional area of a circular coil A =  0.1 cm²
∴ A = 0.1 x 10^-4 m²
(a) Total torque on the coil:
\(\tau=\text { IBA } \cos \theta\)
Here, the plane of the loop is perpendicular to the direction of magnetic field
\(\therefore \theta=90^{\circ} \)
\(\therefore \tau=\text { IBA } \cos 90^{\circ} \)    (∵ cos90^o = 0)
= 3 x 0.2 x 0.1 x 10^-4 x 0
= zero
∴ Total Torque on the coil = zero
(b) Total force on the coil \(F=B q v \sin \theta \)
\(Here\ \theta=0 \)
\(\therefore F=0.2 \times \mathrm{qv} \times \sin \theta=\text { zero } \)
Total force on the coil = zero
c) Average force: F = Bqv
Charge density \(=\sigma=\frac{q}{A} \)
∴ Charge  \(q =\sigma A \)
\(=10^{28} \times 0.1 \times 10^{-4} \)
\(q=10^{23} \mathrm{C} \)
Force = Bll and I = 1 m
F = 0.2 x 3 x 1 = 0.6 N
∴ Average force on each electron
\(=\frac{F}{q}=\frac{0.6}{10^{23}} \)
\(F =0.6 \times 10^{-23} \mathrm{~N} \)

Linear mass density of the conductor is = 0.2 g/m
Mass per unit length \(\frac{M}{l}=0.2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\)
Magnetic field B = 1T.
Acceleration due to gravity, g = 10 ms^-2
Force \(=\frac{m}{l} \times g\)
= 0.2 x 10^-3 x 10 = 0.2 x 10^-2
F = 2 x 10^-3 N   ....(1)
If the coil is placed in the magnetic field then the force acting on the coil is
F= BIl   ....(2)
From the equation (1) and (2) we get
BIl = 2 x 10^-3
∴ 1 x L x I = 2 x 10^-3
∴ I = 2 x 10^-3 A      [∴ l = 1m]
∴ I = 2mA

Magnetic moment of a bar magnet = \({ \vec { { p } } }_{ m }\)

First, When the bar magnet is cut into two pieces along the axis of the magnet. New magnetic pole strength is \(\frac{q_{m}}{2}\), but magnetic length is 2l
Second, Each piece is further cut into two pieces along the axis of the magnet, we get 
New magnetic pole strength is, 
\(q_{m}^{{\prime}}=\frac{q_{m}}{4}\)
But, magnetic length = 2l.

So, new magnetic moment of each piece is
\(\vec{p_m}_{\text {new }} =q'_{m} \times2 l \)
\(=\frac{q_{m}}{4} \times 2 l \)
\(=\frac{1}{4}\left(q_{m} \times 2 l\right) \)
\(\because \overrightarrow{\mathbf{p}}_{\mathrm{m}_{\text {new }}} =\frac{1}{4} (\vec{p_{m})} \)    \((\because{p}_{m} =(q_{\mathrm{m}} \times 2 l) )\)

Hysteresis is the phenomenon of lagging of magnetic induction behind the magnetising field.

Ampere's circuital law states that the line integral of magnetic field over a closed loop is μ₀, times net current enclosed by the loop.
\(\oint _{ c }^{ }{ \vec { B } \vec { dl } } \) = μ₀I _enclosed.

Magnetic permeability is the measure of ability of the material to allow the passage of magnetic field lines through it.

Magnetic susceptibility is defined as the ratio of the intensity of magnetisation (\(\vec { M } \)) induced in the material due to the magnetising field |\(\vec H\)|.
\( \chi _{ m }=\frac { |\vec { M } | }{ |\vec { H } | } \).

Coulomb's inverse square law states that the force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them.
\(\vec { F } =k \frac { { q }_{ m_{A} }{ q }_{ m_{B} } }{ { r }^{ 2 } } \hat { r } \)

Magnetic dipote moment \(\vec { p }_{ m } \) is defined as the product of its pole strength and magnetic length. It is a vector quantity denoted by \(\vec { p }_{ m } \)
\(\vec { p }_{ m } ={ q }_{ m }\vec { d } \)

Magnetic flux is defined as the number of magnetic field lines crossing per unit area kept normal to the direction of lines of force.

(i) When a substance is placed in a uniform magnetising field, the substance gets magnetised.
(ii) The total magnetic field inside the specimen is equal to the sum of the magnetic field produced in vacuum due to the magnetising field and the magnetic field due to the induced magnetism of the substance.

Magnetic field B = 1 T
Mass of the proton, m_p = 1.67 x 10⁻²⁷kg
Charge of the proton, q = 1.60 x 10⁻¹⁹C
\(f=\frac { qB }{ { { 2\pi m }_{ p } } } =\frac { \left( 1.60\times { 10 }^{ -19 } \right) \left( 1 \right) }{ 2\left( 3.14 \right) \left( 1.67\times { 10 }^{ -27 } \right) } \)
= 15.3 x 10⁶ Hz = 15.3 MHz

Electric field, E = 6.0 x 10⁶ N C^-1 and magnetic field, B = 0.83 T.
Then.
\(v=\frac { E }{ B } =\frac { { 6.0\times 10 }^{ 6 } }{ 0.83 } =7.23\times { 10 }^{ 6 }{ ms }^{ -1 }\)
When an electron goes with this velocity, it shows null deflection. Since the accelerating potential is 200 V, the electron acquires kinetic energy because of this accelerating potential. Hence,
\(\frac { 1 }{ 2 } mv^{ 2 }=eV \)
\(v=\sqrt { \frac { 2eV }{ m } }\)
Since the mass of the electron, m = 9.1 x 10⁻³¹kg and charge of an electron, \(\left| q \right| =e=1.6\times { 10 }^{ -19 }C.\) The velocity acquired by the electron due to accelerating potential 200 V is
\({ v }_{ 200 }=\sqrt { \frac { 2\left( 1.6\times { 10 }^{ -19 } \right) \left( 200 \right) }{ \left( 9.1\times { 10 }^{ -31 } \right) } } =8.39\times { 10 }^{ 6 }m{ s }^{ -1 }\)
Since the speed v₂₀₀ > v, the electron is deflected towards direction of Lorentz force. So, in order to have null deflection, the potential, we have to supply is
\(v=\frac { { 1mv }^{ 2 } }{ 2\quad e } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 7.23\times { 10 }^{ 6 } \right) ^{ 2 } }{ 2\times \left( 1.6\times { 10 }^{ -19 } \right) } \)
V = 148.65 V

Velocity of the particle is \(\vec { v } =v\hat { j } \)
(a) Magnetic field is along positive y-direction, this implies \(\vec B=B\hat { j } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { j } )=\vec 0\)
So, no force acts on the particle when it moves along the direction of magnetic field.
(b) Since the magnetic field points in positive z - direction, this implies, \(\vec { B } =B\hat { k } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { k } )=qvB\vec i \)
Therefore, the magnitude of the Lorentz force is qvB and direction is along positive x - direction.
(c) Magnetic field is in zy - plane and making an angle θ with the velocity of the particle, which implies \( {\vec B } =Bcos\theta \hat { j } +Bsin\theta \hat { k } \)

From Lorentz force,
\({ \vec F }_{ m }=q(v\hat { j } )\times (Bcos\theta \hat { j } +Bsin\theta \hat k)\) 
\(=qvBsin\theta \hat { i } \)

Given that I = 1 A and radius r = 1 m
B_straightwire =  \(=\frac { { \mu }_{ ° }I }{ 2\pi r } =\frac { 4\pi \times { 10 }^{ -7 }\times 1 }{ 2\pi \times 1 } =2\times { 10 }^{ -7 }T\)
But the Earth’s magnetic field is B_earth \(\sim { 10 }^{ -5 }T\)
So, B_straightwire is one hundred times smaller than B_Earth.

The magnetic field due to current in the upper semicircle and lower semicircle of the circular coil are equal in magnitude but opposite in direction. Hence, the net magnetic field at the center of the loop (at point O) is zero \(\overset { \rightarrow }{ B } =\overset { \rightarrow }{ 0 } \).

Using right hand rule, current flows upwards.

The slope of M-H graph measures the magnetic susceptibility, which is given by
\({ x }_{ m }=\frac { M }{ H } \)
Material X : Slope is positive and larger value. So, it is a ferromagnetic material.
Material Y : Slope is positive and lesser value than X. So, it could be a paramagnetic material.
Material Z : Slope is negative and hence, it is a diamagnetic material.

The susceptibility of material X is
X_m,x = \(\frac { \left| \overset { \rightarrow }{ M } \right| }{ \left| \overset { \rightarrow }{ H } \right| } =\frac { 500 }{ 1000 } =0.5\)
The susceptibility of material Y is
X_{m,y =} \(\frac { \left| \overset { \rightarrow }{ M } \right| }{ \left| \overset { \rightarrow }{ H } \right| } =\frac { 2000 }{ 1000 } =2\)
Since, susceptibility of material Y is greater than that of material X, material Y can be easily magnetized than X.

\(\overset { \rightarrow }{ B } =\mu _{ ° }(\overset { \rightarrow }{ H+ } \overset { \rightarrow }{ M } )\) 
But from equation (3.33), in vector form,
\(\overset { \rightarrow }{ M } ={ x }_{ m }\overset { \rightarrow }{ H } \)
Hence, \(\overset { \rightarrow }{ B } =\mu _{ ° }({ x }_{ m }+1)\overset { \rightarrow }{ H } \Rightarrow \overset { \rightarrow }{ B } =\mu \overset { \rightarrow }{ H } \)
where, \(\mu =\mu _{ ° }({ x }_{ m }+1)\Rightarrow { x }_{ m }+1=\frac { \mu }{ \mu _{ ° } } =\mu _{ r }\)
\(\Rightarrow { x }_{ m }=\mu _{ r }-1\)

The magnitude of the force between two poles is given by
\( F =k\frac { { q }_{ m_A }{ q }_{ { m }_{ B } } }{ { r }^{ 2 } } \)
(Given : F = 9 × 10^–3 N, r = 10 cm = 10 × 10^–2 m
Since q_mA = q_mB = q_m, we have
9 x 10^-3 = 10^-7 x \(\frac { { q }_{ m }^{ 2 } }{ { \left( 10\times { 10 }^{ -2 } \right) }^{ 2 } } \Rightarrow { q }_{ m }\) = 30NT^-1 

The total flux emanating from the closed surface S enclosing the dipole is zero. So,
\({ \Phi }_{ B }=\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero,
\(\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
This is similar to Gauss’s law in electrostatics.

The geometrical length of the bar magnet is 12 cm
Magnetic length = \(\frac { 5 }{ 6 } \times \)(geometrical length)
= \(\frac { 5 }{ 6 } \times \)12 = 10 cm
In this figure, the dot implies the pole points.

 (a) a bar magnet cut into two pieces along its length:

When the bar magnet is cut along the axis into two pieces, new magnetic pole strength is \({ q }_{ m }^{ ' }=\frac { { q }_{ m } }{ 2 } \) but magnetic length does not change. So, the magnetic moment is
\({ p }_{ m }^{ ' }={ q' }_{ m }2l\)
\({ p }_{ m }^{ ' }=\frac { { q }_{ m } }{ 2 } 2l=\frac { 1 }{ 2 } ({ q }_{ m }2l)=\frac { 1 }{ 2 }p_ m\)
In vector notation, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\)
(b) a bar magnet cut into two pieces perpendicular to the axis:

When the bar magnet is cut perpendicular to the axis into two pieces, magnetic pole strength will not change but magnetic length will be halved. So the magnetic moment is
\({ p }_{ m }^{ ' }={ q }_{ m }\times \frac { 1 }{ 2 } (2l)=\frac { 1 }{ 2 } ({ q }_{ m }.2l)=\frac { 1 }{ 2 } { p }_{ m }\) 
In vector notation, \(\vec{p}'_m=\frac{1}{2}\vec{p}_m\) 

B_H = 0.15 G and B_V = 0.26 G
tan I = \(\frac { 0.26 }{ 0.15 } \Rightarrow I=ta{ n }^{ -1 }(1.732)=60°\)
The resultant magnetic field of the Earth is
\(B=\sqrt { { B }_{ H }^{ 2 }+{ B }_{ V }^{ 2 } } =0.3G\)