The magnetic field at the centre of a square loop due to either arm is
\( \mathrm{B}_{1}=\frac{\mu_{o}}{4 \pi}\left(\frac{I}{L / 2}\right)\left(\sin \phi_{1}+\sin \phi_{2}\right) \)
\(Here, \ \phi_{1}=\phi_{2}=45^{\circ}\)
\(\therefore B_{1} =\frac{\mu_{o}}{4 \pi} \times \frac{2 I}{L}\left[\sin 45^{\circ}+\sin 45^{\circ}\right] \)
\(=\frac{\mu_{o}}{4 \pi} \times \frac{2 I}{L}\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right] \)
\(B_{1} =\frac{\mu_{o}}{4 \pi} \times \frac{2 I}{L} \times\left(\frac{2}{\sqrt{2}}\right)=\frac{\mu_{0}}{4 \pi} \times 2 \frac{\sqrt{2} I}{L} \)
\(\therefore B_{1} =\frac{\mu_{o}}{4 \pi L} \times(2 \sqrt{2}) I \)   .....(1)
Magnetic field at the centre of a square loop carrying a current I
\({B}=4 B_{1}=4 \times \frac{\mu_{o}}{4 \pi} \times \frac{2 \sqrt{2} I}{L}\)
\({B}=\frac{\mu_{o}}{\pi} \times \frac{2 \sqrt{2} I}{L}\)        .........(2)
Substituting I = 1.5 and l = 50 x 10^-2 m in the equations (2) we get
\(B =\frac{4 . \pi \times 10^{-7}}{\pi} \times 2 \sqrt{2} \times \frac{1.5}{50 \times 10^{-2}} \)
\(=8 \sqrt{2} \times 10^{-7} \times 0.03 \times 10^{2} \)
\(=8 \sqrt{2} \times 10^{-7} \times 3 \times 10^{-2} \times 10^{2} \)
\(=24 \sqrt{2} \times 10^{-7} \)
\(=24 \times 1.414 \times 10^{-7}=33.936 \times 10^{-7} \mathrm{~T} \)
\(\mathrm{~B} =3.3936 \times 10^{-6} \mathrm{~T} \)
∴ Magnetic field B = 3.4 x 10^-6 T

M = 100 g = 100 x 10^-3 kg
M = 0.1 kg
g = 10 m/s²
N = 5
B = 0.5 T
R = 20 cm = 20 x 10^-2 m
When, the sphere is in translational equilibrium
\(\mathrm{f}_{\mathrm{s}}-\mathrm{Mg} \sin \theta =0 \ldots(1) \)
\(\mathrm{f}_{\mathrm{s}} =\mathrm{Mg} \sin \theta \)
When, the sphere is in rotational equilibrium,
Torque ፒ = μB sin θ
For Equilibrium, f_sR - ፒ = 0
\(f_{s} R-\mu B \sin \theta=0 \ldots(2)\)
eqn (1) subs. eqn (2)

\(M g \sin \theta R-\mu B \sin \theta =0 \)
\(\sin \theta(M g R-\mu B) =0 \)
\(M g R =\mu B \)
\(\text { Now } \mu =N I \pi R^{2} \)
\(\therefore \operatorname{MgR} =N I \pi R^{2} B \)
\(\therefore I =\frac{M g}{N \pi R B} \)
\(I =\frac{0.1 \times 10}{5 \times \pi \times 20 \times 10^{-2}\times 0.5 }\)
\(=\frac{1}{\pi \times 50 \times 10^{-2}}=\frac{1 \times 10^{2}}{\pi \times 50}=\frac{100}{\pi \times 50}\)
\(\mathbf{I} =2 / \pi \mathrm{A} \)

Uniform magnetic field B = 0.8 T
Angle of orientation θ = 30°
Torque, ፒ = 0.2 Nm.
(i) We know that torque ፒ = P_mB sinθ
\(0.2=p_\mathrm{m} \times 0.8 \times \sin 30^{\circ} \)
\(0.2=p_\mathrm{m} \times 0.8 \times \frac{1}{2} \)
0.2 = 0.4 p_m
\(p_m=\frac{0.2}{0.4}=0.5 \mathrm{Am}^{2}\)
(ii) Work done by the applied force to move the magnet from stable to unstable position.
\(\mathrm{W} =-p_\mathrm{m}B\left[\cos \theta_{2}-\cos \theta_{1}\right] \)
\(\mathrm{W} =-p_\mathrm{m}B\left(\cos 180^{\circ}-\cos 0^{\circ}\right) \)
\(\mathrm{W} =-p_\mathrm{m}B(-1-1)=2 \mathrm{p_mB} \)
\(\mathrm{W} =2 \times 0.5 \times 0.8=\mathbf{0 . 8} \mathbf{J} \)
Work done by the applied magnetic field are in opposite direction
\(\mathbf{W}_{\text {mag }}=-0.8 \mathrm{J}\)

N = 5 turns
A = 2 x 10^-2 m²
B = 4 x 10^-2 Wb m^-2
K = 4 x 10^-9 N m deg^-1
(a) Current sensitivity
\({ I }_{ s }=\frac { NAB }{ K } =\frac { 5\times 2\times { 10 }^{ -2 }\times 4\times { 10 }^{ -2 } }{ 4\times 10^{ -9 } } \)
= 10⁶ divisions per ampere
\(I\mu A=\) 1microambire =10^-6ampere
Therefore,
\({ I }_{ s }={ 10 }^{ 6 }\frac { div }{ A } =1\frac { div }{ { 10 }^{ -6 }A } =1\frac { div }{ \mu A } \)
\({ I }_{ s }=1div{ \left( \mu A \right) }^{ -1 }\)
(b) Voltage sensitivity
\({ V }_{ s }=\frac { \theta }{ V } =\frac { 50div }{ 25mv } =2\times { 10 }^{ 3 }{ div \ V }^{ -1 }\)
(c) The resistance of the galvanometer is
\({ R }_{ g }=\frac { { I }_{ s } }{ { v }_{ s } } =\frac { { 10 }^{ 6 }\frac { div }{ A } }{ { 2\times }10^{ 6 }\frac { div }{ V } } =0.5\times { 10 }^{ 3 }\frac { V }{ A } =0.5k\Omega \)

The linear density of the rod i.e., mass per unit length of the rod is 0.25 kg m^-1
\(\Rightarrow \frac { m }{ 1 } =0.25kg \ { m }^{ -1 }\)
Let I will be the current flowing in the metallic rod. The direction of the electric current is in the paper. The direction of magnetic force IBl is given by Fleming’s left-hand rule.

For equilibrium of the rod,
mg sin 45° = IBI cos45°
\(\Rightarrow 1=\frac { 1 }{ B } \frac { m }{ l } g \tan45°\)
\(=\frac { 0.25kg{ m }^{ -1 } }{ 0.25T } \times 1\times 9.8{ ms }^{ -2 }\)
\(\Rightarrow \) 1 = 9.8A
So, we need to supply current of 9.8 A to keep the metallic rod stationary.

Since isotopes are singly ionized, they have equal charge which is equal to the charge of an electron, q = - 1.6 x 10^-19 C. Mass of uranium \(_{ 92 }^{ 235 }{ U and _{ 92 }^{ 238 }{ U } }\) are 3.90 x 10^-25 kg and 3.95 x 10^-25 kg respectively. Magnetic field applied, B = 0.500 T. Velocity of the electron is 1.00 x 10⁵ m s^-1, then 
(a) the radius of the path of \(_{ 92 }^{ 235 }{ U }\) is r₂₃₅
\({ r }_{ 235 }=\frac { { m }_{ 235 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =48.8\times { 10 }^{ -2 }m\)
r₂₃₅ = 48.8cm 
The diameter of the semi-circle due to \(_{ 92 }^{ 235 }{ U\ \ is \ \ { d }_{ 235 }=2{ r }_{ 235 } }\) = 97.6 cm
The radius of the path of _{\(_{ 92 }^{ 238 }{ U\ is\ 2{ r }_{ 238 }\ then}\)}
\({ r }_{ 238 }=\frac { { m }_{ 238 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =49.4\times { 10 }^{ -2 }m\)
r₂₃₈ = 49.4 cm
The diameter of the semi-circle due to  \(^{ 238 }_{92}{ U\ is \ 2{ r }_{ 238 } \ =98.8 \ cm}\)
Therefore the separation distance between the isotopes is \(\triangle d={ d }_{ 238 }-{ d }_{ 235 }=1.2 \ cm\)
(b) The time taken by each isotope to complete one semi-circular path are
\({ t }_{ 235 }=\frac { \text{ magnitude of the displacement} }{ velocity } \)
\(=\frac { 97.6\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.76\times { 10 }^{ -6 }s=9.76\mu s\)
\({ t }_{ 238 }=\frac { \text{magnitude of the displacement }}{ velocity } \)
\(=\frac { 98.8\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.88\times { 10 }^{ -6 }s=9.88\mu s\) 

Magnetic field \(\overset { \rightarrow }{ B } ={ 0.500\hat { i } T } \)
Velocity of the particle
\(\hat { v } \) = (1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) ms^-1
Charge of the proton q = 1.67 x 10^-19 C
Mass of the proton m = 1.67 x 10^-27kg
(a) The force experienced by the proton is \(\overset { \rightarrow }{ F } \) = q(\(\overset { \rightarrow }{ v } \) x \(\overset { \rightarrow }{ B } \) )
= 160 x 10^-19 x  ((1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) x (0.500 \(\hat { i } \)))
\(\overset { \rightarrow }{ F } \)= 1.60 x 10^-14 \(\hat { j } \)  N
Therefore, from Newton’s second law,
\(\overset { \rightarrow }{ a } =\frac { 1 }{ m } \overset { \rightarrow }{ F } =\frac { 1 }{ 1.67\times { 10 }^{ -27 } } (1.60\times { 10 }^{ -14 })\hat j\)
\(=9.58\times { 10 }^{ 12 }\hat jm{ s }^{ -2 }\)
(b) Trajectory is helical Radius of helical path is
\(R=\frac { { mv }_{ z } }{ \left| q \right| B } =\frac { 1.67\times { 10 }^{ -27 }\times 2.00\times { 10 }^{ 5 } }{ 1.60\times { 10 }^{ -19 }\times 0.500 } \)
= 4.175 x 10^-3m = 4.18mm
Pitch of the helix is the distance travelled along x-axis in a time T, which is P = v_x T
But time, \(T=\frac { 2\pi }{ \omega } =\frac { 2\pi m }{ \left| q \right| B } =\frac { 2\times 3.14\times 1.67\times { 10 }^{ -27 } }{ 1.60\times 1{ 0 }^{ -19 }\times 0.500 } =13.1\times { 10 }^{ -8 }s\)
Hence, pitch of the helix is
\(p={ v }_{ x }T=(1.95\times { 10 }^{ 5 })(13.1\times { 10 }^{ -8 })=25.5\times { 10 }^{ -3 }m=25.5mm\)
The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

For a charged particle moving on a magnetic field, \(\vec { F } \)= q(\(\vec { v } \)x\(\vec { B } \))
The work done by the magnetic field is
\(W=\int { \overset { \rightarrow }{ F } .{ d \vec r } =\int { \overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } dt } } \)
\(W=q\int { \left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) } .\overset { \rightarrow }{ v } dt=0\)
Since \(\overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \) is perpendicular to \(\vec { v } \) and hence \(\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) .\overset { \rightarrow }{ v } =\overset { \rightarrow }{ 0 } \) 
This means that Lorentz force does no work on the particle. From work-kinetic energy theorem, (Refer section 4.2.6, XI th standard Volume I)
\(\frac { dw }{ dt } =p=0\)
Since \(\overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } =0\Rightarrow \overset { \rightarrow }{ F } \ and \ \overset { \rightarrow }{ v } \) are perpendicular to each other. The angle between Lorentz force and velocity of the charged particle is 90^o. Thus Lorentz force changes the direction of the velocity but not the magnitude of the velocity. Hence Lorentz force does no work and also does not alter kinetic energy of the particle. 

The magnetic field of a solenoid (inside) is
\({ B }_{ L,N }=\mu _{ ° }\frac { NI }{ L } \)
(a) length of the solenoid becomes twice and fixed number of turns
L → 2L (length becomes twice)
N → N (number of turns remains constatnt)
The magnetic field is
\({ B }_{ L,N }=\mu _{ ° }\frac { NI }{ 2L } =\frac { 1 }{ 2 } { B }_{ L,N }\)
(b) both the length of the solenoid and number of turns are doubled
L → 2L (length becomes twice)
N → 2N (number of turns becomes twice)
The magnetic field is
\({ B }_{2 L,2N }=\mu _{ ° }\frac { 2NI }{ 2L } ={ B }_{ L,N }\)
(c) the number of turns becomes twice but the fixed length of the solenoid
L → L (length is fixed)
N → 2N (number of turns becomes twice)
The magnetic field is
\({ B }_{ L,2N }=\mu _{ ° }\frac { 2NI }{ L } ={ 2B }_{ L,N }\)
From the above results,
\({ B }_{ L,2N }>{ B }_{ 2L,2N }>{ B }_{ 2L,N }\)
Thus, strength of the magnetic field is increased when we pack more loops into the same length for a given current.

Density of the magnet is
Density = \(\frac { Mass }{ volume } \Rightarrow Volume=\frac { Mass }{ Density } \)
\(Volume=\frac { 200\times 1{ 0 }^{ -3 }kg }{ \left( 8\times 1{ 0 }^{ -3 }kg \right) \times 1{ 0 }^{ 6 }{ m }^{ -3 } } =25\times { 10 }^{ -6 }{ m }^{ 3 }\)
Magnitude of magnetic moment p_m = 2A m²
Intensity of magnetization,
\(I=\frac { magnetic\ moment }{ Volume } =\frac { 2 }{ 25\times { 10 }^{ -6 } } \)
M = 0.8 x 10⁵ Am^-1

The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T
Deflection is 
\(\theta =60°\Rightarrow tan60°=\sqrt { 3 } =1.732\)
\(I=\frac { 2R{ B }_{ H } }{ { \mu }_{ ° }N } tan\theta \)
\(=\frac { 2\times 0.12\times 25\times 1{ 0 }^{ -6 } }{ 4\times 1{ 0 }^{ -7 }\times 3.14\times 100 } \times 1.732=0.82\times 1{ 0 }^{ -1 }A\)
I = 0.082 A

Let  \(\vec{p}_m\)be the dipole and before switching ON the external magnetic field, there is no orientation. Therefore, the energy U = 0.
As soon as external magnetic field is switched ON, the magnetic dipole orient parallel (θ = 0^o) to the magnetic field with energy,
U_parallel = U_minimum = -p_mBcos 0
U_parallel = -p_mB
since cos 0^o = 1
Otherwise, the magnetic dipole orients anti-parallel (θ = 180^o) to the magnetic field with energy,
U _{anti-parallel} = U _maximum = -p_mBcos 180
\(\Rightarrow \) U_{anti-parallel} = P_mB
since cos 180^o = -1

Given magnetic moment 0.5 J T^-1 and distance r = 0.1 m
(a) When the point lies on the axial line of the bar magnet, the magnetic field for short magnet is given by
\({ { \vec B }_{ axial } } =\frac { { \mu }_{ ° } }{ 4\pi } \left( \frac { 2{ p }_{ m } }{ { r }^{ 3 } } \right) \hat { i } \)
\({ { \vec B }_{ axial } } =1{ 0 }^{ -7 }\times \left( \frac { 2\times 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) =1\times { 10 }^{ -4 }\hat { i } \ T\)
Hence, the magnitude of the magnetic field along axial is B_axial = 1 x 10^-4 T and direction is towards South to North.
(b) When the point lies on the normal bisector (equatorial) line of the bar magnet, the magnetic field for short magnet is given by
\({ {\vec B }_{ equatorial } } =-\frac { { \mu }_{ ° } }{ 4\pi } \frac { { p }_{ m } }{ { r }^{ 3 } } \hat { i } \)
\({ {\vec B }_{ equatorial } } =-1{ 0 }^{ -7 }\left( \frac { 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) \hat { i } =-0.5\times 1{ 0 }^{ -4 }\hat { i } \ T \)
Hence, the magnitude of the magnetic field along axial is B_equatorial = 0.5 x 10^-4 T and direction is towards North to South.
Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.

(i) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
(ii) A galvanometer is converted into a volmeter by connecting high resistance R_s in series with the galvanometer.

Biot-Savart's law states that, the magnitude of magnetic field \(d\vec { B } \) at a point P at a distance of r from the small elemental length taken on a conductor carrying current varies
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length of element \(\vec { dl } \)
(iii) directly as the sine of the angle θ between \(\vec { dl } \) and \(\hat { r } \).
(iv) inversely as the square of the distance r between the point P and length of element \(\vec { dl } \).
\(d\vec { B } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I\vec { dl } \times \hat { r } }{ { r }^{ 2 } } \)

Lorentz deduced the force experienced by a moving charge in the magnetic field \(\overrightarrow{F_{m}}\)
\(\overrightarrow{F_{m}}=q(\vec{v} \times \vec{B})\)   ........(1)
In magnitude, 
\(\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin \theta\)   ...(2)
The equations (1) and (2) imply
i) \(\overrightarrow{F_{m}}\) directly proportional to the magnetic field \(\overrightarrow{B}\)
ii) \(\overrightarrow{F_{m}}\) is directly proportional to the velocity \(\overrightarrow{v}\) of the moving charge
iii) \(\overrightarrow{F_{m}}\) is directly proportional to sine of the angle between the velocity and magnetic field
iv) \(\overrightarrow{F_{m}}\) is directly proportional to the magnitude of the charge q
v) The direction of \(\overrightarrow{F_{m}}\) is always perpendicular to \(\overrightarrow{v}\) and \(\overrightarrow{B}\) as \(\overrightarrow{F_{m}}\) is the cross product of \(\overrightarrow{v}\) and \(\overrightarrow{B}\)
vi) The direction of \(\overrightarrow{F_{m}}\) on negative charge is opposite to the direction of \(\overrightarrow{F_{m}}\) on positive charge provided other factors are identical.
vii) If velocity \(\overrightarrow{v}\) of the charge q is along magnetic field \(\overrightarrow{B}\) then, \(\overrightarrow{F_{m}}\) is zero.