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CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 4

MOVING CHARGES AND MAGNETRISM

Magnetic field

Oersted’s observation

Ampere’s swimming rule (for direction of deflection of compass needle)

Magnetic field due to a current element - Biot Savart’s law

Magnetic field lines

Magnetic field due to a straight current carrying conductor

Maxwell’s cork screw rule

Right hand thumb rule for straight current carrying wire

Strength of magnetic field due to straight current carrying conductor

Magnetic field on the axis of a current carrying circular coil

Magnetic field at centre of the coil

Right hand thumb rule for a circular current carrying loop

Ampere’s circuital law

Magnetic field outside due to current carrying wire of infinite length by Ampere’s circuital law

Magnetic field inside a current carrying wire by Ampere’s circuital law

Solenoid

Magnetic field outside a solenoid is practically zero

Magnetic field inside a solenoid

Toroid

Magnetic force on a moving charge

Magnetic force on a current carrying conductor

Fleming’s left hand rule

Motion of charge in a magnetic field

Helical motion of charged particle

Motion of charge in combined electric and magnetic fields

Velocity selector

Cyclotron

Structure of cyclotron

Principle of cyclotron

Maximum kinetic energy gained by the particle in the cyclotron

Limitations of the cyclotron

Force between two infinitely long parallel current carrying conductors

Definition of Ampere

Defnition of Coulomb

Torque acting on a current carrying coil placed inside a uniform magnetic field

Current carrying loop as a magnetic dipole

The clock face rule

Difference between electric dipole and magnetic dipole

Magnetic dipole moment of a revolving electron

Moving coil galvanometer

Converting galvanometer to ammeter - Shunt resistance

Current sensitivity of galvanometer

Converting galvanometer to voltmeter

Voltage sensitivity of galvanometer

CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 4

MOVING CHARGES AND MAGNETRISM

Magnetic Field

The space in the surrounding of a magnet or any current carrying conductor in which its magnetic influence can be experienced.

Oersted’s Observation

A magnetic field is produced in the surrounding of any current carrying conductor. The direction of this magnetic field can be obtained by Ampere’s swimming rule.

SI unit of magnetic field is Wm-2 or T (telsa).

The strength of magnetic field is one Tesla, if a charge of one coulomb, when moving with a velocity of 1 ms-1 along a direction perpendicular to the direction of the magnetic field experiences a force of one Newton.

CGS unit of magnetic field is called Gauss or Oersted.

Ampere’s Swimming Rule (For direction of deflection of compass needle)

If a man is swimming along the wire in the direction of current his face turned towards the needle, so that the current enters through his feet, then northpole of the magnetic needle will be deflected towards his left hand.

Image result for ampere's swimming rule

The conventional sign for a magnetic field coming out of the plane and normal to it is a dot i.e., ‘.’, the magnetic field perpendicular to the plane in the downward direction is denoted by cross, i.e., ‘×’.

Magnetic field due to a current element - Biot Savart’s law

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The magnetic field produced by a current carrying element of length dl, carrying current I at a point separated by a distance r is given by

dB =   μoI4πdl×rr3

Or

dB =   μo4πI dl sin θr2  

where,

θ is the angle between the direction of the current and r

μo is absolute permeability of the free space.

The direction of magnetic field dB is that of dl×r.

Value of

 μo4π=10-7  Tm/A

ɛoμo=4πɛo μo4π= 19×109×10-7 m-2

Comparision of Electrostatic and Magnetic Force

Magnetic Field Lines

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Magnetic Field Due to a Straight Current Carrying Conductor

The magnetic field lines due to a straight current carrying conductor are concentric circles having centre on the conductor and in a plane perpendicular to the conductor.

Maxwell’s Cork Screw Rule

If a right handed cork screw is imagined to be rotated in such a direction that tip of the screw points in the direction of the current, then direction of rotation of thumb gives the direction of magnetic field lines.

Image result

Right hand thumb rule for straight current carrying wire

To determine the direction of the magnetic field due to a long wire, the right hand thumb rule is,

Grasp the wire in your right hand with your extended thumb pointing in the direction of the current. Your fingers will curl around in the direction of the magnetic field.

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Strength of magnetic field due to straight current carrying conductor

B =μo4π Ir(sinϕ1+sinϕ2)

where ϕ1 and ϕ2 are angles, which the lines joining the two ends of the conductor to the observation point make with the perpendicular from the observation point to the conductor.

Proof

Let us consider a straight conductor carrying current I. We need to find the magnetic field at a point P, which is at a perpendicular distance r from the conductor. Point P makes angles ϕ1 and ϕ2 with the perpendicular to the conductor.

Now let us consider a current element on the conductor. The line joining it with P makes an angle θ with the perpendicular.

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From the geometry of the figure, we can see that,

l = r tan θ 

Differentiating this equation we get,

dl = r sec2⁡ θ 

PQ = r cos θ 

Now, as per Biot Savart’s law, the field at P due to this element, will be

dB =   μo4πI dl sin π2-θPQ2=   μo4πI dl cos θ  PQ2

Putting the value of dl and PQ, we get,

dB =   μo4πI r sec2θ   × cos θr2cos θ2

=   μoI 4πrcos θ d 

Integrating the above equation, we get,

0BdB=   μoI 4πr-ϕ1ϕ2cos θ  

B=   μoI 4πrsin-ϕ1ϕ2

B =μo4π Ir(sinϕ2-sin(-ϕ1))

B =μo4π Ir(sinϕ1+sinϕ2)

Magnetic field on the axis of a current carrying circular coil

Consider a point P on the axis of the current carrying coil, at a distance x from the centre of the coil.

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From the geometry,

The angle between r and dl is 90°.

Hence field due to current carrying element dl is

dB =  μo4π I dl sin 90°r2= μo4π I dl r2

The direction of magnetic field is perpendicular to the plane containing r and dl . So the magnetic field dB has two components,

Similarly, consider another current carrying element dl', which is diametrically opposite to the point A. The magnetic field due to this current carrying element, dB' also has two components

Since dB cos θ and dB’ cos θ are equal in magnitude and opposite in direction, they cancel each other.

On the other hand, the components dB sin θ and dB’ sin θ are equal in magnitude and in same direction so they add up.

Hence total magnetic field due to the circular current carrying coil at the axis is,

B =dBsin= 02πaμo4πI dl r2ar 

= 02πaμo4πI dl r2×ar   

Now,

sin=ar 

and

r = a2+x2

Therefore,

B = 02πaμo4πI a dla2+x232

=μo4πI a (a2+x2)32 02πadl

=μo4π×I a a2+x232×2πa

 B=μo2I a2 (a2+x2)32

Magnetic field at centre of the coil,

x = 0  B=μo2I  a

If there are ‘n’ number of turns in the circular coil, then the magnetic field is

B=μo2n I  a

Right hand thumb rule for current carrying loop

Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field.

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Ampere’s circuital law

The line integral of magnetic field induction B around any closed path in vacuum is equal to μ0 times the total current threading the closed path (or passing through the surface enclosed by the closed path), i.e.,

B.dl=B dl = μoI2πr×2πr= μoI

where B is the magnetic field, dl is small length element, μo is the absolute permeability of free space and I is the current.

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Magnetic field due to current carrying wire of infinite length using Ampere’s circuital law

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Magnetic field outside a current carrying wire

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Let us consider a cylindrical conductor of radius ‘a’, carrying current I.

Take an Amperian loop of radius r > a.

The current passing through the circular cross section of radius r is I

Now from Amperes’ circuit law, we have,

02πRB.dl= μoI 

B × 2πr = μoI 

 B =μoI2πr

Therefore, magnetic field outside the wire B = μoI2πr

Magnetic field inside a current carrying wire

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Let us consider a cylindrical conductor of radius ‘a’, carrying current I.

Take an Amperian loop of radius r < a.

The current density through the conductor

J=Iπa2

Hence current passing through the circular cross section of radius r

I1= πr2×Iπa2=Ir2a2

Now from Amperes’ circuit law, we have,

02πrB.dl= μo I1 

B × 2πr = μo I r2a2

 B =μo I r2a2×12πr

Therefore, magnetic field inside the wire B = μo I r2πa2

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SOLENOID

A solenoid is a closely wound helix of insulated copper wire. It behaves like magnet when current passes through it.

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A long solenoid means that its length is much larger compared to its radius.

Each turn can be regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns.

The magnetic field outside the solenoid is practically zero.

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We have two "rows" of currents, the top one (current coming out of the page) and the bottom one (current going into the page). We follow the "right hand thumb rule", and take into account the fact that the fields from all of the individual loops add together.

Inside the solenoid, the fields from the two rows point in the same direction, reinforcing each other. This results in a net field that is strong.

Outside the solenoid, the fields due to the circular loops between neighbouring turns tend to cancel each other.

The top row produces a combined magnetic field that points (mostly) left above itself, and (mostly) right below itself.

The field due to bottom row points (mostly) right above itself, and (mostly) left below itself.

The fields at any point outside the solenoid, from the two rows, point in opposite directions, mostly canceling each other (totally canceling for an infinite solenoid).

Therefore the magnetic field outside a solenoid is weak or almost zero.

The magnetic field inside the solenoid

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Let us consider a rectangular Amperian loop abcd as shown in the diagram.

The field is zero along cd (outside the solenoid).

The field component is zero along transverse sections bc and ad,. Thus, these two sections make no contribution.

Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h.

Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I nh, where I is the current in the solenoid.

From Ampere’s circuital law

Magnetic field at a point on one end of a long solenoid is given by

B =μo n I2

TOROID

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A toroidal solenoid is an anchor ring around which large number of turns of a copper wire are wrapped. A toroid is an endless solenoid in the form of a ring.

The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Let us consider three circular Amperian loops 1, 2 and 3.

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By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop

Loop 1 encloses no current, hence B1 = 0

For loop 3, the current coming out of the plane of the paper is cancelled exactly by the current going into it. Hence B3 at Q is also zero.

For loop 2, the current enclosed Ie is (for N turns of toroidal coil) N I, that is

B (2πr) = μoNI

B =μoNI2πr

If we take,

 n =N2πr N =2πrn,

we get,

Therefore, the field inside the toroid is the same as the field inside the solenoid.

Magnetic field inside a toroid is constant and is always tangential to the circular closed path.

Magnetic field at any point inside the empty space surrounded by the toroid and outside the toroid, is zero, because net current enclosed by this space is zero.

MAGNETIC FORCE ON A MOVING CHARGE

A point charge q (moving with a velocity v and, located at r at a given time t) in presence of both the electric field E(r) and the magnetic field B(r), experiences force due to both electrical and magnetic field. The force on an electric charge q due to both electrical and magnetic field is called the Lorentz force and can be written as

FE =qE

And

FB =q v× B 

Total force, F =FE + FB=q[E+v× B]

The magnetic force has following characteristics,

  1. It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge.

  2. The magnetic force

    FB = q  v× B= qvB sin θ n̂

    includes a vector product of velocity and magnetic field. The vector product makes the force due to magnetic field zero if velocity and magnetic field are parallel or anti-parallel and maximum when they are at right angles to each other.

    The force acts in a (sideways) direction perpendicular to both of them and Its direction is given by the screw rule or right hand rule for vector (or cross) product.

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  1. Only a moving charge feels the magnetic force. The magnetic force is zero if charge is not moving (as |v|= 0).

Magnetic force on a current carrying conductor

Consider a rod of a uniform cross-sectional area A and length L. Let the number density of mobile charge carriers (electrons) in it be n. Then the total number of mobile charge carriers in it is nLA. For a steady current I in this conducting rod, each mobile carrier has an average drift velocity vd.

In the presence of an external magnetic field B, the force on these carriers is:

FB= (nLA)q vd × B

where q is the value of the charge on a carrier.

Now,

Current density j = n q vd

and

Current I = |(n q vd)| A

Thus,

F = n q vd L A×  B 

In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod.

For any arbitrary shape of the conductor,

F = Σ I  dL ×  B

This summation can be converted to an integral in most cases.

Fleming’s left hand rule

[Given direction of external magnetic field and current, find direction of force on it]

If we stretch the thumb, the forefinger and the central finger of left hand in such a way that all three are perpendicular to each other, then, if Forefinger represents the direction of magnetic Field, Central finger represents the direction of Current flowing through the conductor, then thuMb will represent the direction of magnetic force (or Motion).

Motion of charge in a magnetic field – circular motion

A force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle.

In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).

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The force due to an electric field, i.e., qE, can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.

Consider motion of a charged particle in a uniform magnetic field.

Let us first take the case of v perpendicular to B. The perpendicular force, qvB, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field.

Now since θ = 90o, If r is the radius of circular path, then,

mv2r= qvB 

The radius of the circular path

 r =mv Bq 

And

v=qBrm 

The larger the momentum (or velocity) the larger is the radius and bigger the circle described. If ω is the angular frequency, then

v = ωr

And

ω = 2π ν = qBm

Time period,

T=2πω=2πmBq

When a charged particle enters the magnetic field at any angle except 90°, it follows helical path.

In this case, the velocity has a component along B. This component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion.

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The distance travelled by the charged particle in one time period due to component of velocity v cos θ (denoted by v||), is called pitch of the path

p = v|| T  

=T × v cos θ = 2πm v||q B

Motion of charge in combined electric and magnetic fields

Velocity selector

A charge q moving with velocity v in presence of both electric and magnetic fields experiences a force given by

F =[ qE + qv × B]  FE + FB

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We have,

FE=qE=qEĵ  

and

 FB = qv × B

Therefore total force,

F=qE-vBĵ

Thus, electric and magnetic forces are in opposite directions. If we adjust the value of E and B such that magnitudes of the two forces are equal, then, total force on the charge is zero and the charge will move in the fields undeflected.

Or

qE = qvB 

v =EB

We can therefore select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass).

The crossed E and B fields, therefore, serve as a velocity selector.

Only particles with speed EB pass undeflected through the region of crossed fields.

CYCLOTRON

Cyclotron is a device used to accelerate positively charged particles such as proton, deuteron etc.

Structure of cyclotron

Cyclotron consists of two large metallic D’s, connected to an oscillator which can change polarity of the D’s at regular intervals, a source of positively charged particles at the centre and a deflector plate as shown in the diagram.

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Principle of Cyclotron

A positively charged particle can be accelerated through a moderate electric field by crossing it again and again by use of strong magnetic field.

The frequency νa of the applied voltage is adjusted so that the polarity of the Dees is reversed in the same time that it takes the ions to complete one half of the revolution.

The requirement νa = νc is called the resonance condition.

The phase of the supply is adjusted so that when the positive ions arrive at the edge of D1, D2 is at a lower potential and the ions are accelerated across the gap.

Inside the Dees the particles travel in a region free of the electric field. The increase in their kinetic energy is qV each time they cross from one Dee to another (V refers to the voltage across the dees at that time).

The radius of their path goes on increasing each time their kinetic energy increases. The ions are repeatedly accelerated across the Dees until they have the required energy to have a radius approximately that of the Dees.

They are then deflected by a magnetic field and leave the system via an exit slit.

Radius of circular path

 r =mvBq

Cyclotron frequency

νc =Bq2πm

 T=2πmBq

where m and q are mass and charge of the positive ion and B is strength of the magnetic field.

Maximum kinetic energy gained by the particle in the cyclotron

Now,

v=qBrom

Hence,

Emax=12mv2=B2q2ro22m 

where, ro = maximum radius of circular path.

Where,

Limitations of Cyclotron

  1. Cyclotron cannot accelerate uncharged particle like neutron.

  2. The positively charged particles having large mass i.e., ions cannot move at limitless speed in a cyclotron

Force between two infinitely long parallel current carrying conductors

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Consider two long parallel conductors ‘a’ and ‘b’ separated by a distance d and carrying (parallel) currents Ia and Ib, respectively. The conductor ‘a’ produces, the same magnetic field Ba at all points along the conductor ‘b’ which is downwards.

Ba =μo2π Iad

The conductor ‘b’ carrying a current Ib will experience a sideways force due to the field Ba. The direction of this force is towards the conductor ‘a’.

Fba = Ib L Ba= μo2π IaIb dL

Similarly

Fab = Ia L Bb= -μo2π IaIb dL

We can see that Fab = - Fba

The force is attraction force if current in both conductors is in same direction and repulsion if current in both conductors is in opposite direction.

Definition of Ampere

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2×10–7 newtons per metre of length.

Defnition of Coulomb

When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its cross-section in 1s is one coulomb (1C).

Torque acting on a current carrying coil placed inside a uniform magnetic field

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Consider the rectangular loop placed in such a way that the uniform magnetic field B is in the plane of the loop.

The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop,

F1 = I b B

Similarly it exerts a force F2 on the arm CD which is directed out of the plane of the paper,

F2 = I b B F1

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Thus, the net force on the loop is zero. However there is a torque on the loop due to the pair of forces F1 and F2.

τ = F1×a2+ F2×a2

=IbBa sin = I ab Bsin 

Torque acting on a current carrying coil with N turns,

Where,

Current loop as magnetic dipole

The current carrying loop behaves as a small magnetic dipole placed along the axis. One face of the loop behaves as north-pole while the other face of loop behaves as south-pole.

The clock face rule

While looking at the face of the coil, if the current is found to be flowing in the anti-clockwise direction, then the face of the coil will behave like the north pole and if the current is found to be flowing in the clockwise direction, the face of the coil will behave like south pole.

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Each magnetic dipole has some magnetic moment (m). The vector m is given by

m = N I A

Hence

τ = m×B

This expression is similar to electrostatic case

τ = p×E

SI unit of m are Am2

Stable Equilibrium

When m and B are parallel.

Unstable Eqilibrium

When m and B are anti-parallel.

Difference between electric dipole and magnetic dipole

Electrical Dipole is combination of two elementary charges, Charges can exist individually. However no magnetic monopoles exist.

The magnetic dipole moment of a revolving electron

Ampere suggested that all magnetism is due to circulating currents. This seems to be partly true and no magnetic monopoles have been seen so far. However, elementary particles such as an electron or a proton also carry an intrinsic magnetic moment, not accounted by circulating currents.

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Let us consider an electron that is revolving around in a circle of radius r with a velocity v.

The electron of charge (– e) (e = + 1.6 × 10–19 C) performs uniform circular motion around a stationary heavy nucleus of charge +Ze.

Current I = eT

T is the time period of revolution. Let r be the orbital radius of the electron, and v the orbital speed, then,

T =2πrv

I=ev2πr

The magnetic moment, denoted by μl, associated with this circulating current is,

 μl = Iπr2 =evr2

Multiplying and dividing the right-hand side of the above expression by the electron mass me, we have,

μl = emevr2 me=el2me, 

Where l = angular momentum of the electron

Moving coil galvanometer

It is a device used for the detection and measurement of the currents. It works on the principle of conversion of electrical energy into mechanical energy. When a current flows through the coil, a torque acts on it.

In equilibrium, deflecting torque = restoring torque, or

k = NIAB      =NABkI

Hence θ ∝ I

where,

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Q: Why Galvanometer cannot be used to measure current?

  1. Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of μA.

  2. For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit.

Ammeter - Shunt resistance

To overcome these difficulties, a small resistance rs, called shunt resistance, is attached in parallel with the galvanometer coil; so that most of the current passes through the shunt.

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Since, RG >> rs, the resistance of this arrangement is,

RG rs RG +  rs  rs

Current sensitivity of galvanometer

The deflection produced per unit current in galvanometer is called its current sensitivity. Current sensitivity

Is =θI=NBAk

Voltmeter

The galvanometer can be used as a voltmeter. For this it must be connected in parallel with that section of the circuit. It must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large. Hence a large resistance R is connected in series with the galvanometer.

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The voltmeter resistance is,

RG + R R (very large)

Voltage sensitivity of galvanometer

The deflection produced per unit voltage applied across the ends of galvanometer is called its voltage sensitivity.

Vs =θV=NBAkR

For a sensitive galvanometer

  1. N should be large

  2. B should be large

  3. A should be large

  4. k should be small

Q: Increasing the current sensitivity may not necessarily increase the voltage sensitivity. Explain.

If N → 2N, i.e., we double the number of turns, then current sensitivity will double.

However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire (Number of turns).