**NCERT solutions for class 12 Physics chapter 4 Moving charges and magnetism: **A moving charge in a magnetic field can produce a force and a current-carrying conductor in a uniform magnetic field will experience a force. The solutions of NCERT class 12 physics chapter 4 moving charges and magnetism covers problems based on these two basic concepts. NCERT solutions will help you to boost the concepts studied in the chapter. Two main laws discussed in the unit are Ampers law and Biort-savart law. Based on these laws many problems are discussed in CBSE NCERT solutions for class 12 physics chapter 4 moving charges and magnetism. The galvanometer that is discussed in chapter 3 current electricity is discussed in detail with its working principle here. NCERT solutions for class 12 physics chapter 4 moving charges and magnetism helps students for solving homework problems.

The right-hand thumb rule, Flemings right-hand rule and Flemings left-hand rule are three important concepts required for solutions of NCERT class 12 physics chapter 4 moving charges and magnetism. The problems related to finding the directions can be solved using these three rules or by using the concepts of vectors. The NCERT solutions for class 12 physics chapter 4 moving charges and magnetism also helps in preparing competitive exams like NEET and JEE Mains.

The magnitude of the magnetic field at the centre of a circular coil of radius r carrying current I is given by,

For 100 turns, the magnitude of the magnetic field will be,

(current=0.4A, radius = 0.08m, permeability of free space = 410^{-7 }TmA^{-1})

The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,

In this case

(current=35A, distance= 0.2m, permeability of free space = 410^{-7 }TmA^{-1})

The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,

In this case

(current=50A, distance= 2.5m, permeability of free space = 410^{-7 }TmA^{-1})

The current is going from the North to South direction in the horizontal plane and the point lies to the East of the wire. Applying Maxwell's right-hand thumb rule we can see that the direction of the magnetic field will be vertically upwards.

The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,

In this case (current=35A, distance= 0.2m, permeability of free space = 410^{-7 }TmA^{-1})

The current in the overhead power line is going from the East to West direction and the point lies below the power line. Applying Maxwell's right-hand thumb rule we can see that the direction of the magnetic field will be towards the South.

The magnetic force on an infinitesimal current-carrying conductor in a magnetic field is given by where the direction of vector dl is in the direction of flow of current.

For a straight wire of length l in a uniform magnetic field, the Force equals to

In the given case the magnitude of force per unit length is equal to

|F| = 0.158sin30^{o} (I=8A, B=0.15 T, =30^{o})

=0.6 Nm^{-1}

For a straight wire of length l in a uniform magnetic field, the Force equals to

In the given case the magnitude of the force is equal to

|F| = 0.27100.03sin90^{o} (I=10A, B=0.27 T, =90^{o})

=0.081 N

The direction of this force depends on the orientation of the coil and the current-carrying wire and can be known using the Flemings Left-hand rule.

The magnitude of magnetic field at a distance r from a long straight wire carrying current I is given by,

In this case the magnetic field at a distance of 4.0 cm from wire B will be

(I=5 A, r=4.0 cm)

The force on a straight wire of length l carrying current I in a uniform magnetic field B is given by

, where is the angle between the direction of flow of current and the magnetic field.

The force on a 10 cm section of wire A will be

(B=2.5 T, I=8 A, l = 10 cm, =90^{o})

The magnitude of the magnetic field at the centre of a solenoid of length l, total turns N and carrying current I is given by

, where is the permeability of free space.

In the given question N= number of layers of windingnumber of turns per each winding

N=5400=2000

I=8.0 A

l=80 cm

The magnitude of torque experienced by a current-carrying coil in a magnetic field is given by

where n = number of turns, I is the current in the coil, A is the area of the coil and is the angle between the magnetic field and the vector normal to the plane of the coil.

In the given question n = 20, B=0.8 T, A=0.10.1=0.01 m^{2}, I=12 A, =30^{o}

=0.96 Nm

The coil, therefore, experiences a torque of magnitude 0.96 Nm.

**10 (a) **Two moving coil meters, and have the following particulars:

The torque experienced by the moving coil M_{1} for a current I passing through it will be equal to

The coil will experience a restoring torque proportional to the twist

The current sensitivity is therefore

Similarly, for the coil M_{2, }current sensitivity is

Their ratio of current sensitivity of coil M_{2} to that of coil M_{1} is, therefore,

**10.(b) **Two moving coil meters, and have the following particulars:

(The spring constants are identical for the two meters).

Determine the ratio of voltage sensitivity of and

The torque experienced by the moving coil M_{1} for a current I passing through it will be equal to

The coil will experience a restoring torque proportional to the twist

we know V=IR

Therefore,

Voltage sensitivity of coil M_{1 }=

Similarly for coil M_{2} Voltage sensitivity =

Their ratio of voltage sensitivity of coil M_{2} to that of coil M_{1}

The magnetic force on a moving charged particle in a magnetic field is given by

Since the velocity of the shot electron is perpendicular to the magnetic field, there is no component of velocity along the magnetic field and therefore the only force on the electron will be due to the magnetic field and will be acting as a centripetal force causing the electron to move in a circular path. (if the initial velocity of the electron had a component along the direction of the magnetic field it would have moved in a helical path)

Magnetic field(B)=

Speed of electron(v)=

Charge of electron=

Mass of electron=

The angle between the direction of velocity and the magnetic field = 90^{o}

Since the force due to the magnetic field is the only force acting on the particle,

In exercise 4.11 we saw

Time taken in covering the circular path once(time period (T))=

Frequency,

From the above equation, we can see that this frequency is independent of the speed of the electron.

**13 (a) **A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Number of turns in the coil(n)=30

The radius of the circular coil(r)=8.0 cm

Current flowing through the coil=6.0 A

Strength of magnetic field=1.0 T

The angle between the field lines and the normal of the coil=60^{o}

The magnitude of the counter-torque that must be applied to prevent the coil from turning would be equal to the magnitude of the torque acting on the coil due to the magnetic field.

=3.13 Nm

A torque of magnitude 3.13 Nm must be applied to prevent the coil from turning.

From the relation we can see that the torque acting on the coil depends only on the area and not its shape, therefore, the answer won't change if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area.

**NCERT solutions for moving charges and magnetism additional exercise:**

Using the right-hand thumb rule we can see that the direction of the magnetic field due to coil X will be towards the east direction and that due to coil Y will be in the West direction.

We know the magnetic field at the centre of a circular loop of radius r carrying current I is given by

(towards East)

(towards west)

The net magnetic field at the centre of the coils,

B_{net}=B_{y }- B_{x}

=1.5710^{-3} T

The direction of the magnetic field at the centre of the coils is towards the west direction.

Strength of the magnetic field required is

nI

Therefore keeping the number of turns per unit length and the value of current within the prescribed limits such that their product is approximately 8000 we can produce the required magnetic field.

e.g. n=800 and I=10 A.

Show that this reduces to the familiar result for the field at the centre of the coil.

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

For finding the field at the centre of coil we put x=0 and get the familiar result

Let a point P be at a distance of l from the midpoint of the centres of the coils.

The distance of this point from the centre of one coil would be R/2+l and that from the other would be R/2-l.

The magnetic field at P due to one of the coils would be

.

The magnetic field at P due to the other coil would be

Since the direction of current in both the coils is same the magnetic fields B_{1 }and B_{2} due to them at point P would be in the same direction

B_{net }=B_{1}+B_{2}

Since l<<R we can ignore term l^{2}/R^{2}

Since the above value is independent of l for small values it is proved that about the midpoint the Magnetic field is uniform.

Outside the toroid, the magnetic field will be zero.

The magnetic field inside the core of a toroid is given by

Total number of turns(N)=3500

Current flowing in toroid =11 A

Length of the toroid, l=

(r_{1}=inner radius=25 cm, r_{2}=outer radius=26 cm)

The magnetic field in the empty space surrounded by the toroid is zero.

The charged particle is not deflected by the magnetic field even while having a non zero velocity, therefore, its initial velocity must be either parallel or anti-parallel to the magnetic field i.e. It's velocity is either towards the east or the west direction.

Yes, its final speed will be equal to the initial speed if it has not undergone any collision as the work done by the magnetic field on a charged particle is always zero because it acts perpendicular to the velocity of the particle.

The electron would experience an electrostatic force towards the north direction, therefore, to nullify its force due to the magnetic field must be acting on the electron towards the south direction. By using Fleming's left-hand rule we can see that the force will be in the north direction if the magnetic field is in the vertically downward direction.

Explanation:

The electron is moving towards the east and has a negative charge therefore is towards the west direction, Force will be towards south direction if the magnetic field is in the vertically downward direction as

(a) The electron has been accelerated through a potential difference of 2.0 kV.

Therefore K.E of electron =

Since the electron initially has velocity perpendicular to the magnetic field it will move in a circular path.

The magnetic field acts as a centripetal force. Therefore,

The electron has been accelerated through a potential difference of 2.0 kV.

Therefore K.E of electron = 1.610^{-19}2000=3.210^{-16} J

The component of velocity perpendicular to the magnetic field is

The electron will move in a helical path of radius r given by the relation,

r=5m

r=510^{-4 }m

r=0.5 mm

The component of velocity along the magnetic field is

The electron will move in a helical path of pitch p given by the relation,

p=5.4510^{-3} m

p=5.45 mm

The electron will, therefore, move in a helical path of radius 5 mm and pitch 5.45 mm.

(i)

Let the beam consist of particles having charge q and mass m.

After being accelerated through a potential difference V its velocity can be found out by using the following relation,

(ii)

Using the value of v from equation (ii) in (i) we have

In order for the tension in the wires to be zero the force due to the magnetic field must be equal to the gravitational force on the rod.

mass of rod=0.06 g

length of rod=0.45m

the current flowing through the rod=5 A

A magnetic field of strength 0.261 T should be set up normal to the conductor in order that the tension in the wires is zero

**Answer:**

If the direction of the current is reversed the magnetic force would act in the same direction as that of gravity.

Total tension in wires(T)=Gravitational force on rod + Magnetic force on rod

The total tension in the wires will be 1.176 N.

Since the distance between the wires is much smaller than the length of the wires we can calculate the Force per unit length on the wires using the following relation.

Current in both wires=300 A

Distance between the wires=1.5 cm

Permeability of free space=410^{-7} TmA^{-1}

F=1.2 Nm^{-1}

The length of wire inside the magnetic field is equal to the diameter of the cylindrical region=20.0 cm=0.2 m.

Magnetic field strenth=1.5 T.

Current flowing through the wire=7.0 A

The angle between the direction of the current and magnetic field=90^{o}

Force on a wire in a magnetic field is calculated by relation,

F=2.1 N

This force due to the magnetic field inside the cylindrical region acts on the wire in the vertically downward direction.

Magnetic field strenth=1.5 T.

Current flowing through the wire=7.0 A

The angle between the direction of the current and magnetic field=45^{o}

The radius of the cylindrical region=10.0 cm

The length of wire inside the magnetic field,

Force on a wire in a magnetic field is calculated by relation,

F=2.1 N

This force due to the magnetic field inside the cylindrical region acts on the wire in the vertically downward direction.

This force will be independent of the angle between the wire and the magnetic field as we can see in the above case.

Note: There is one case in which the force will be zero and that will happen when the wire is kept along the axis of the cylindrical region.

The wire is lowered by a distance d=6cm.

In this case, the length of the wire inside the cylindrical region decreases.

Let this length be l.

F=1.68 N

This force acts in the vertically downward direction.

The magnetic field is

Current in the loop=12 A

Area of the loop = lengthbreadth

A=0.10.05

A=0.005 m^{2}

The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y direction. The force on the loop is zero.

The magnetic field is

Current in the loop=12 A

Area of the loop = lengthbreadth

A=0.10.05

A=0.005 m^{2}

(same as that in the last case)

The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y direction. The force on the loop is zero. This was exactly the case in 24. (a) as well.

The magnetic field is

Current in the loop=12 A

Area of the loop = lengthbreadth

A=0.10.05

A=0.005 m^{2}

The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-x-direction. The force on the loop is zero.

The magnetic field is

Current in the loop=12 A

Area of the loop = lengthbreadth

A=0.10.05

A=0.005 m^{2}

The torque on the loop has a magnitude of 0.018 Nm and at an angle of 240^{o} from the positive x-direction. The force on the loop is zero.

The magnetic field is

Current in the loop=12 A

Area of the loop = lengthbreadth

A=0.10.05

A=0.005 m^{2}

Since the area vector is along the direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.

The magnetic field is

Current in the loop=12 A

Area of the loop = lengthbreadth

A=0.10.05

A=0.005 m^{2}

Since the area vector is in the opposite direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.

The force on the loop in all the above cases is zero as the magnetic field is uniform

As we know the torque on a current-carrying loop in a magnetic field is given by the following relation

It is clear that the torque, in this case, will be 0 as the area vector is along the magnetic field only.

The total force on the coil will be zero as the magnetic field is uniform.

The average force on each electron in the coil due to the magnetic field will be eV_{d}B where V_{d} is the drift velocity of the electrons.

The current is given by

where n is the free electron density and A is the cross-sectional area.

The average force on each electron is

The magnetic field inside the solenoid is given by

n is number of turns per unit length

n=1500 m^{-1}

Current in the wire I_{w }= 6 A

Mass of the wire m = 2.5 g

Length of the wire l = 2 cm

The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances weight of the wire

Therefore a current of 108.37 A in the solenoid would support the wire.

The galvanometer can be converted into a voltmeter by connecting an appropriate resistor of resistance R in series with it.

At the full-scale deflection current(I) of 3 mA the voltmeter must measure a Voltage of 18 V.

The resistance of the galvanometer coil G =

The galvanometer can be converted into a voltmeter by connecting a resistor of resistance in series with it.

The galvanometer can be converted into an ammeter by connecting an appropriate resistor of resistance R in series with it.

At the full-scale deflection current(I) of 4 mA, the ammeter must measure a current of 6 A.

The resistance of the galvanometer coil is G =

Since the resistor and galvanometer coil are connected in parallel the potential difference is the same across them.

The galvanometer can be converted into an ammeter by connecting a resistor of resistance in parallel with it.

As CBSE board exam is concerned, the solutions of NCERT class 12 physics chapter 4 moving charges and magnetism is important. In 2019 CBSE board exam 12 % of questions are asked from chapter 4 and 5. Same questions discussed in the NCERT solutions for class 12 physics chapter 4 moving charges and magnetism can be expected in the board exams.

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