(a) From Flemings left-hand rule, the electron deflects in an anticlockwise direction.
As the electron comes out of the magnetic field region, it will describe a semi-circular path.
Magnetic force provides a centripetal force
(b) Be v = \(\frac { mv^{ 2 } }{ r } \) (or) Be = \(\frac { mv }{ r } \)
Time is taken T= \(\frac { \pi r }{ v } =\frac { \pi m }{ Be } \)
T = \(\frac { 3.14\times 9.1\times 10^{ -31 } }{ 1.6\times 10^{ -19 }\times 10^{ -5 } } \)
= \(\frac { 3.14\times 9.1\times 10^{ -7 } }{ 1.6 } \)
T = 1.97 x 10^-7S

The cyclotron's frequency v = 18 μHz
= 8 x 10⁶ Hz
The mass of the proton m = 1.67 x 10^-27 kg
The charge of the proton q = 1.6 x 10^-19C
Radius of the dees r = 50 cm = 50 x 10^-2 m
Magnetic field B = ?
k.E of the proton k.E = ?
Magnetic field B = \(\frac { 2\pi mv }{ q } \)
B = \(\frac { 2\times 3.14\times 1.67\times 10^{ -27 }\times 8\times 10^{ 6 } }{ 1.6\times 10^{ -19 } } \)
= \(\frac { 83.90\times 10^{ -21 } }{ 1.6\times 10^{ -19 } } \)
= 52.438 x 10^-2
B = 0.524T
k.E = \(\frac { 1 }{ 2 } \) mv²
v = rω = r x 2πv
= 0.5 x 2 x 3.14 x 8 x 10⁶
= 25.12 x 10⁶ m/s
k.E = \(\frac { 1 }{ 2 } \) x 1.67 x 10^-27 x (25.12 x 10⁶)²
=\(\frac { 41.95\times 10^{ -21 }\times 25.12\times 10^{ 6 } }{ 2 } \)
= 526.89 x 10^-15 J (or) 5.269 x 10^-13J
To convert J into MeV
= \(\frac { 526.89\times 10^{ -15 } }{ 1.6\times 10^{ -19 } } \)  = 3.29 x 10⁶ eV
k.E = 3.29 x 10⁶ MeV

Current I₁ = 2A ; I₂ = 5A
Two wires are separated by a distance a = 1 cm
= 1 x 10^-2m
The force between two parallel wires per unit length
F =?
F = \(\frac { { \mu }_{ 0 } }{ 2\pi } .\frac { { I }_{ 1 }{ I }_{ 2 } }{ a } \)
= 2 x 10^-7 x \(\frac { 2\times 5 }{ 1\times { 10 }^{ -2 } } \)
F = 20 x 10^-5 N
This force is repulsive F = 20 x 10^-5 N.

Magnetic moment μ = 0.6 J/T
Angle of inclination with magnetic field θ = 30°
Torque acts on bar magnet ፒ = 0.051 J
Magnitude of magnetic field B = ?
ፒ = \(\vec { \mu } \times \vec { B } \) = μB sinθ
B = \(\frac { \tau }{ \mu sin\theta } =\frac { 0.051 }{ 0.6\times sin{ 30 }^{ 0 } } \)
= \(\frac { 0.051 }{ 0.6\times \frac { 1 }{ 2 } } =\frac { 2\times 0.051 }{ 0.6 } =\frac { 0.102 }{ 0.6 } \)
B = 0.17T
(ii) The position of maximum energy corresponds to a position of stable equilibrium.
The energy (v) = -mB cosθ
when θ = 0°, v = -μB = minimum energy Hence when the bar magnet is placed parallel to the magnetic field, it is the state of stable equilibrium.

In the general case, the unit normal vector \(\hat { n } \) and magnetic field \(\vec { B } \) is with an angle 8 as shown in Figure.

(a) The force on section PQ
\(\vec { i } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { { F }_{ PQ } } =\vec { Il } \times \vec { B } =IaB(\hat { j } \times \hat { j } )=-IaB\hat { k } \)
Since the unit vector normal to the plane \(\hat { n } \) is along the direction of .\(\vec { k } \)
(b) The force on section QR
\(\vec { l } =bcos\left( \frac { \pi }{ 2 } -\theta \right) \hat { i } -sin\left( \frac { \pi }{ 2 } -\theta \right) \hat { k } \)
\(\vec { { F }_{ QR } } =\vec { Il } \times \vec { B } =-IbB\left( \frac { \pi }{ 2 } -\theta \right) \hat { j } \)
\(\vec { { F }_{ QR } } =-IbBcos\theta \hat { j } \)
(c) The force on section RS
\(\vec { l } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { { F }_{ RS } } =\vec { Il\times \vec { B } } =IaB(\hat { j } \times \hat { j } )=-IaB\hat { k } \)
Since the unit vector normal to the plane is along the direction of \(\hat { k } \).
(d) The force on section SP
\(\vec { l } =bcos\left( \frac { \pi }{ 2 } -\theta \right) \hat { i } +sin\left( \frac { \pi }{ 2 } +\theta \right) \hat { k } \quad \vec { B } =B\hat { i } \)
\(\vec { { F }_{ SP } } =\vec { Il } \times \vec { B } =IbBsin\left( \frac { \pi }{ 2 } -\theta \right) \hat { j } \)
\(\vec { { F }_{ SP } } =-IbBcos\theta \hat { j } \)
The net force on the rectangular loop is
\(\vec { { F }_{ net } } =\vec { { F }_{ PQ } } +\vec { { F }_{ QR } } +\vec { { F }_{ RS } } +\vec { { F }_{ SP } } \)
\({ F }_{ net }=IaB\hat { k } -IbBcos\theta \hat { j } -IaB\hat { k } +IbBcos\theta \hat { j } \)
\(\vec { { F }_{ net } } =\vec { 0 } \)
Hence, the net force on the rectangular loop in this configuration is also zero. Notice that the force on section QR and SP is not zero here. But, they have equal and opposite effects, but we assume that the loop to be rigid, so no deformation. So, no torque was produced by these two sections.
Even though the forces PQ and RS also are equal and opposite, they are not collinear. So these two forces constitute a couple as shown in Figure (a). Hence the net torque produced by these two forces about the axis of the rectangular loop is given by
\(\vec { { \tau }_{ net } } =baBIsin\theta \hat { k } =ABIsin\theta \hat { k } \)

\(\vec { OA } =\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (-\hat { i } )+\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (-\hat { k } )\)
=\(\frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } )\)
\(\vec { OB } =\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (\hat { i } )+\frac { b }{ 2 } cos\left( \frac { \pi }{ 2 } -\theta \right) (\hat { k } )\)
=\(\frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } )\)
\(\vec { OA } \times \vec { { F }_{ PQ } } =\left\{ \frac { b }{ 2 } (-sin\theta \hat { i } +cos\theta \hat { k } \right\} \times \left\{ IaB\hat { k } \right\} \)
= \(\frac { 1 }{ 2 } IabBsin\theta \hat { j } \)
\(\vec { OA } \times \vec { { F }_{ RS } } =\left\{ \frac { b }{ 2 } (sin\theta \hat { i } +cos\theta \hat { k } \right\} \times \left\{ -IaB\hat { k } \right\} \)
= \(\frac { 1 }{ 2 } IabBsin\theta \hat { j } \)
The net torque \(\vec { \tau _{ net } } =IaBsin\theta \hat { j } \)         ..........(1)
Note that the net torque is in the positive y-direction which tends to rotate the loop in a clockwise direction about the y axis. If the current is passed in the other way (P⟶S⟶R⟶Q⟶P), then total torque will point in the negative y-direction which tends to rotate the loop in an anticlockwise direction about the y-axis.
Another important point is to note that the torque is less in this case compared to the earlier case (where the \(\hat { n } \) is perpendicular to the magnetic field \(\vec { B } \)). It is because the perpendicular distance is reduced between the forces \(\vec { { F }_{ PQ } } \) and \(\vec { { F }_{ RS } } \) in this case.
The equation (1) can also be rewritten in terms of magnetic dipole moment \(\vec { { p }_{ m } } =I\vec { A } =Iab\hat { n } \)
\(\vec { \tau _{ net } } =\vec { p } \times \vec { B } \).

Consider a single rectangular loop PQRS kept in a uniform magnetic field \(\vec { B } \). Let a and b be the length and breadth of the rectangular loop respectively. Let \(\hat { n } \) be the unit vector normal to the plane of the current loop. This unit vector \(\hat { n } \) completely describes the orientation of the loop. Let \(\vec { B } \) be directed from north pole to south pole of the magnet as shown in Figure.

When an electric current is sent through the loop, the net force acting is zero but there will be net torque acting on it. For the sake of understanding, we shall consider two configurations of the loop;
(i) unit vector \(\hat { n } \) points perpendicular to the field
(ii) unit vector points at an angle θ with the field.
when unit vector \(\hat { n } \) is perpendicular to the field
In the simple configuration, the unit vector \(\hat { n } \) is perpendicular to the field and plane of the loop is lying on the XY plane as shown in Figure. Let the loop be divided into four sections PQ, QR, RS, and SP. The Lorentz force on each loop can be calculated as follows:
(a) Force on section PQ,
(a) \(\vec { i } =-a\hat { j } \) and \(\vec { B } =B\vec { i } \)
\(\vec { { F }_{ PQ } } =\vec { Il } \times \vec { B } =-IaB(\hat { J } \times \hat { i } )=laB\hat { k } \)
Since the unit vector normal to the plane \(\hat { n } \) is along the direction of \(\hat { k } \).
(b) The force on section QR
\(\vec { l } =\vec { bi } \) and \(\vec { B } =B\hat { i } \)
\(\vec { F_{ QR } } =\vec { Il } \times \vec { B } =-IbB(\hat { j } \times \hat { i } )=\vec { 0 } \)
(c) The force on section RS
\(\vec { l } =a\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { F_{ RS } } =\vec { Il } \times \vec { B } =IaB(\hat { j } \times \hat { i } )=-IaB\hat { k } \)
Since the unit vector normal to the plane is along the direction of -\(\hat { k } \).
(d) The force on section SP
\(\vec { l } =b\hat { j } \) and \(\vec { B } =B\hat { i } \)
\(\vec { F_{ SP } } =\vec { Il } \times \vec { B } =IaB(\hat { i } \times \hat { i } )=\vec { 0 } \)
The net force on the rectangular loop is
\(\vec { { F }_{ net } } =\vec { F_{ PQ } } +\vec { { F }_{ QR } } +\vec { { F }_{ RS } } +\vec { { F }_{ SP } } \)
\(\vec { { F }_{ net } } =IaB\hat { k } +\vec { 0 } -IaB\hat { k } +\vec { 0 } \Rightarrow \vec { { F }_{ net } } =\vec { 0 } \)
Hence, the net force on the rectangular loop in this configuration is zero. Now let us calculate the net torque due to these forces about an axis passing through the center
\(\vec { { \tau }_{ net } } =\overset { 4 }{ \underset { i=1 }{ \Sigma } } \vec { { \tau }_{ i } } =\overset { 4 }{ \underset { i=1 }{ \Sigma } } \vec { { r }_{ i } } \times \vec { { F }_{ i } } \)
=\(\left( \frac { b }{ 2 } IaB+0+\frac { b }{ 2 } IaB+0 \right) \hat { j } \)
\(\vec { { \tau }_{ net } } =abIB\hat { j } \)
Since A = ab is the area of the rectangular loop PQRS, therefore, the net torque for this configuration is
\(\vec { { \tau }_{ net } } =ABI\hat { j } \).

(i) Consider a charged particle of charge 'q' having mass m enters into a region of a uniform magnetic field \(\vec { B } \) with velocity \(\vec { v } \).
(ii) Such that velocity is perpendicular to the magnetic field and velocity \(\vec{v}\).
(iii) The charged particle moves in a circular orbit.
(iv) Lorentz force.

\(\vec { F } =q(\vec { v } \times \vec { B } )\)
In magnitude F = qVB
(v) This Lorentz force acts as centripetal force for the particle to execute circular motion. Therefore,
qvB = m\(\frac { { v }^{ 2 } }{ r } \)
The radius of the circular path is
r = \(\frac { mv }{ qB } =\frac { p }{ qB } \)               ..........(1)
(vi) where p = mv is the magnitude of the linear momentum of the particle. Let T be the time taken by the particle to finish one complete circular motion, then
T = \(\frac { 2\pi r }{ v } \)                  .............(2)
Hence substituting (1) in (2), we get,
T = \(\frac { 2\pi m }{ qB } \)        .............(3)
(vii) Equation (3) is called the cyclotron period. The reciprocal of time period is the frequency f, which is
f = \(\frac { 1 }{ T } \)
f = \(\frac { qB }{ 2\pi m } \)                  ...........(4)
In terms of angular frequency ω,
ω = 2πf = \(\frac { q }{ m } \)B          ...........(5)
(viii) Equations (4) and equation (5) are called cyclotron frequency or gyrofrequency.
(ix) Time period and frequency depend only on charge-to-mass ratio (specific charge) and independent of velocity or radius.

A solenoid is bent in a way that both their ends are joined together to form a closed ring shape, is called a toroid. The magnetic field has constant magnitude inside the toroid whereas, in the interior region (say, at point P) and exterior region (say, at point Q), the magnetic field is zero.
(i) Open space interior to the toroid:
Let us calculate the magnetic field B_p, at point P. We construct an Amperian loop 1 of radius r₁ around the point P as shown in Figure. For simplicity, we take a circular loop so that the length of the loop is its circumference.

L₁ = 2πr₁
Ampere's circuital law for the loop 1 is
\(\int _{ loop1 }^{ }{ \vec { B_{ p } } .\vec { dl } } \) = μ0 Ienclosed
Since, the loop1 encloses no current,
Ienclosed = 0
\(\int _{ loop2 }^{ }{ \vec { B_{ p } } .\vec { dl } } \)  = 0
This is possible only if the magnetic field at point P vanishes i.e.
\(\vec { B_{ p } } \) = 0
(b) In open space exterior to the toroid
Let us calculate the magnetic field B_Q at point Q. We construct an Amperian loop 3 of radius r₃ around the point Q as shown in Figure. The length of the loop is, \(\mathrm{L}_{3}=2 \pi r_{3}\)
Ampere's circuital law for the loop 3 is
\(\oint_{\text {loop } 1} \vec{B}_{Q} \cdot \vec{d} l=\mu_{o} I_{\text {endosed }}\)
Since, in each turn of the toroid loop, current coming out of the plane of paper is cancelled by the current going into the plane of paper. Thus, I_enclosed = 0
\(\oint_{\text {loop } 3} \vec{B}_{Q} \cdot \overrightarrow{d l}=0\)
This is possible only if the magnetic field at point Q vanishes i.e.,
\(\vec{B}_{Q}=0\)
(iii) Inside the toroid :
Let us calculate the magnetic field Bs at point S by constructing an Amperian loop 2 of radius r2 around the point S as shown in Figure. The length of the loop is,  L₂ = 2πr₂
Ampere's circuital law for loop 2 is
\(\int _{ loop2 }^{ }{ \vec { B_{ s } } .\vec { dl } } \) = μ₀ I_enclosed
Let I be the current passing through the toroid and N be the number of turns of the toroid, then
I_enclosed = NI
and \(\int _{ loop2 }^{ }{ \vec { B_{ s } } .\vec { dl } } =\int _{ loop2 }^{ }{ Bdl } cos\theta \)  = B2πr₂
\(\int _{ loop2 }^{ }{ \vec { B_{ s } } .\vec { dl } } \) = μ₀ NI
B_s = \(\frac { NI }{ 2\pi { r }_{ 2 } } \)
The number of turns per unit length is n = \(\frac { N }{ 2\pi { r }_{ 2 } } \), then the magnetic field at point S is,
B_s = μ₀nI

(i) Electron revolves around a nucleus in a circular orbit of radius R.
(ii) Circulating electron is like a current in a circular loop.

\(\vec { { \mu }_{ L } } =I\vec { A } \) ......(1)
In magnitude,
μ_L = IA
If T is the time period of an electron, the current due to the circular motion of the electron is
I = \(\frac { -e }{ T } \) ....(2)
where -e is the charge of an electron. If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit, then
T = \(\frac { 2\pi R }{ v } \)   ..(3)
Using equation (2) and equation (3) in equation (1), we get
μ_L = \(\frac { e }{ \frac { 2\pi R }{ v } } \pi { R }^{ 2 }=\frac { evR }{ 2 } \)   ....(4)
where A = πR² is the area of the circular loop. By definition, the angular momentum of the electron about O is
\(\vec { L } =\vec { r } \times \vec { p } \)
In magnitude
L = Rp = mvR  ...(5)
Using equation (4) and equation (5), we get
\(\frac { { \mu }_{ L } }{ L } =-\frac { \frac { evR }{ 2 } }{ mvR } =\frac { e }{ 2m } \Rightarrow \vec { { \mu }_{ L } } =\frac { e }{ 2m } \vec { L } \) ....(6)
The negative sign indicates that the magnetic moment and angular momentum are in opposite directions.
In magnitude
\(\frac{\mu_L}{L}=\frac{e}{2 m}=\frac{1.60 \times 10^{-19}}{2 \times 9.11 \times 10^{-31}}=0.0878 \times 10^{12} \mathrm{C} \mathrm{kg}^{-1}\)
\(\frac{\mu_L}{L}=8.78 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}=\text { constant }\)
The ratio \(\frac{\mu_L}{L}\) is aconstant known as Gyro-magnetic ratio \(\left(\frac{e}{2 m}\right)\).
According to Bohr quantization
\(\mathrm{L}=\mathrm{nh} / 2 \pi\)
\(\mu_L=\left(\frac{e}{2 m}\right) \mathrm{L}=\frac{\pi e h}{4 \pi m} \)
On substiting known values
\(\mu_L=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2\)
The minimum magnetic moment can be obtained by substituting n = 1
\(\mu_2=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2=9.27 \times 10^{-24} \mathrm{~J} \mathrm{~T}^{-1}\)
where, \(\mu_B=\frac{c h}{4 \pi m}=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2\) is called Bohr magneton. which is used to measure atomic magnetic moments.

When a bar magnet (magnetic dipole) of dipole moment \(\vec { { p }_{ m } } \) is held at an angle θ with the direction of a uniform magnetic field \(\vec { { B } } \), as shown in Figure the magnitude of the torque acting on the dipole is

\(|\vec { \tau _{ B } } |=|\vec { { p }_{ m } } ||\vec { B } |sin\theta \)
If the dipole is rotated through a very small angular displacement dθ against the torque ፒ_B at constant angular velocity, then the work done by external torque \((\vec { { \tau }_{ ext } } )\) for this small angular displacement is given by
dW = \(|\vec { { { \tau }_{ ext } } } |\) dθ
Since the bar magnet to be moved at constant angular velocity, it implies \(|\vec { { \tau }_{ B } } |=|\vec { \tau _{ ext } } |\)
dW = P_mB sinθ dθ
Total work done in rotating the dipole from θ' to θ is
W =\(\int _{ \theta ' }^{ \theta }{ \tau d\theta } =\int _{ \theta ' }^{ \theta }{ p_{ m } } Bsin\theta d\theta ={ p }_{ m }B[-cos\theta d\theta ]_{ \theta ' }^{ \theta }\)
W = p_mB(cosθ - cosθ')
This work done is stored as potential energy in bar magnet at an angle θ when it is rotated from θ' to θ and it can be written as
U = p_mB(cosθ - cosθ')           .........(1)
In fact, equation (1) gives the difference in potential energy between the angular positions θ' and θ. We can choose the reference point θ' = 90°, so that second term in the equation becomes zero and the equation (1) can be written as
U = -p_mB(cosθ)              ............(2)
The potential energy stored in a bar magnet in a uniform magnetic field is given by
U = -\(\vec { { p }_{ m } } .\vec { B } \)                     ............(3)
Case 1
(i) If θ = 0°, then
U = P_mB (cos0⁰) = - P_mB
(ii) If θ = 180°, then
U = P_mB (cos 180°) = p_mB
We can infer from the above two results, the potential energy of the bar magnet is minimum when it is aligned along the external magnetic field and maximum when the bar magnet is aligned anti-parallel to an external magnetic field.

Let the currents flowing through wires XY and PQ be I₁ and I₂
Magnetic field along PQ is \(\mathrm{B}_{1}=\frac{\mu_{o} I_{2}}{2 \pi r}\)
Force per unit length on PQ is \(\frac{F_{2}}{l}=\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)
If the wire XY is slightly displaced and released, it executes simple harmonic motion with the condition that acceleration is directly proportional to the displacement y
\(\therefore a=-\omega^{2} y\)  .....(1) 
The distance between two wires = d
Time period
\(T=\frac{2 \pi}{\omega} \)
\(a=\frac{g}{d} y \) .....(2)
By comparing the equations (1) and (2) we get
\(\omega^{2} =-\frac{g}{d} \ \therefore \omega=\sqrt{\frac{g}{d}} \)
\(\text { Time period } =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{d}{g}} \)
\(\therefore T =2 \pi \sqrt{\frac{d}{g}} \)

Let the current flowing be I.
Let the number of turns per unit length of the solenoid be n
Magnetic field B \(\mathrm{B}=\frac{\mu_{o} N l I}{l}=0 \)
But \(\mathbf{n}=\frac{N}{l} \)
\(B=\mu_{o} n I \)
But the magpetic field at any point on the axis of the solenoid is
\(B=\frac{1}{2}\left(\mu_{o} n I\right)\)
On the top of the solenoid, at a point magnetic field is
\(\mathrm{B}_{1}=\frac{1}{2} \mu_{o} n I \cos \theta_{1}\)
On the bottom of the solenoid, the magnetic field at a point is
\(B_{2} =\frac{1}{2} \mu_{0} n I \cos \theta_{2} \)
Resultant magnetic field \(B =B_{1}-B_{2} \)
\(B =\frac{1}{2} \mu_{0} n I\left[\cos \theta_{1}-\cos \theta_{2}\right] \)

Principle : When a current carrying loop is placed in a uniform magnetic field it experiences a torque.
Construction : A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil as shown in Figure. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet.

The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. In a fine suspension strip, a small plane mirror is attached in order to measure the deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head. In order to pass electric current through the galvanometer, the suspension strip and the spring S are connected to terminals.
Working : Consider a single turn of the rectangular coil PQRS whose length be l and breadth b. PQ = RS = l and QR = SP = b.
Let I be the electric current flowing through the rectangular coil PQRS as shown in Figure. The horse-shoe magnet has hemi - spherical magnetic poles which produces a radial magnetic field. Due to this radial field, the sides QR and SP are always parallel to the magnetic field B and experience no force. The sides PQ and RS are always parallel to the magnetic field and experience force in opposite directions. Due to this, torque is produced.

For single turn, the deflection torque is,
て = bF = bBIl = (lb)BI
て = ABI
since, area of the coil A = lb
For coil with N turns, we get
て = NABI               ........(1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the moment of restoring couple is proportional to the amount of twist θ. Thus
て = Kθ      ............(2)
where K is the restoring couple per unit twist or torsional constant of the spring.
At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equations (1) and (2), we get,
NABI = Kθ
⇒ I =\(\frac { K }{ NAB } \) θ      ...........(3)
(or) I = Gθ
where G = \(\frac { K }{ NAB } \) is called galvanometer constant or current reduction factor of the galvanometer.
Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galvanometer we use are pointer type moving coil galvanometer.

When the current is passed through the loop, the torque is produced. If the torque acting on the loop is increased then the loop will start to rotate. The loop will start to lift if and only if the magnitude of magnetic torque due to current applied equals to the gravitational torque as shown in Figure
T_magnetic = T_{gravitational}
IAB = mgR
\(But\quad { p }_{ m }=IA=I(\pi { R }^{ 2 })\)
\(IR^{ 2 }B=mgR\)
\(\Rightarrow \frac { mg }{ \pi { R }B } \)
The current estimate using this equation should be applied so that one edge of loop is lifted off the surface.