# NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

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.

## 1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

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

$|B|=\frac{\mu _{0}I}{2r}$

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

$|B|=100\times \frac{\mu _{0}I}{2r}$

$|B|=100\times \frac{4\pi \times 10^{-7}\times 0.4}{2\times 0.08}$        (current=0.4A, radius = 0.08m, permeability of free space = 4$\pi \times$10-7 TmA-1)

$=3.14\times10^{-4}T$

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

$|B|=\frac{\mu _{0}I}{2\pi r}$

In this case

$|B|=\frac{4\pi \times 10^{-7}\times 35}{2\pi\times 0.2}=3.5\times 10^{-5}T$            (current=35A, distance= 0.2m, permeability of free space = 4$\pi \times$10-7 TmA-1)

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

$|B|=\frac{\mu _{0}I}{2\pi r}$

In this case

$|B|=\frac{4\pi \times 10^{-7}\times 50}{2\pi\times 2.5}$            (current=50A, distance= 2.5m, permeability of free space = 4$\pi \times$10-7 TmA-1)

$\\=4\times10^{-6}T$

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,

$|B|=\frac{\mu _{0}I}{2\pi r}$

In this case  (current=35A, distance= 0.2m, permeability of free space = 4$\pi \times$10-7 TmA-1)

$|B|=\frac{4\pi \times 10^{-7}\times 90}{2\pi\times 1.5}=1.2\times10^{-5} T$

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.

## 5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of $30 \degree$with the direction of a uniform magnetic field of 0.15 T?

The magnetic force on an infinitesimal current-carrying conductor in a magnetic field is given by $\vec{dF}=I\vec{dl}\times \vec{B}$ 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

$\\\vec{F}=\int_{0}^{l}I\vec{dl}\times \vec{B}\\ |\vec{F}|=BIlsin\theta$

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

|F| = 0.15$\times$8$\times$sin30o         (I=8A, B=0.15 T, $\theta$=30o)

=0.6 Nm-1

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

$\\\vec{F}=\int_{0}^{l}I\vec{dl}\times \vec{B}\\ |\vec{F}|=BIlsin\theta$

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

|F| = 0.27$\times$10$\times$0.03$\times$sin90o         (I=10A, B=0.27 T, $\theta$=90o)

=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.

## 7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

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

$|B|=\frac{\mu _{0}I}{2\pi r}$

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

$|B|=\frac{4\pi \times 10^{-7}\times 5}{2\pi\times 0.04}$             (I=5 A, r=4.0 cm)

$\\=2.5\times10^{-5} T$

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

$F=BIlsin\theta$, where $\theta$ 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

$F=2.5\times10^{-5}\times8\times0.1 \times sin90^\circ$       (B=2.5 T, I=8 A, l = 10 cm, $\theta$=90o)

$F=2\times10^{-5} N$

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

$B=\frac{\mu _{o}NI}{l}$ , where $\mu _{o}$ is the permeability of free space.

In the given question N= number of layers of winding$\times$number of turns per each winding

N=5$\times$400=2000

I=8.0 A

l=80 cm

$B=\frac{4\pi \times 10^{-7}\times 2000\times 8}{0.8}$

$\\B=2.51\times10^{-2} T$

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

$\tau =nBIAsin\theta$

where n = number of turns, I is the current in the coil, A is the area of the coil and $\theta$ 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=0.1$\times$0.1=0.01 m2, I=12 A, $\theta$=30o

$\tau =20\times 0.8\times 12\times 0.01\times sin30^{o}$

=0.96 Nm

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

The torque experienced by the moving coil M1 for a current I passing through it will be equal to $\tau =B_{1}A_{1}N_{1}I$

The coil will experience a restoring torque proportional to the twist $\phi$

$\phi k=B_{1}A_{1}N_{1}I$

The current sensitivity is therefore $\frac{B_{1}A_{1}N_{1}}{k}$

Similarly, for the coil M2, current sensitivity is $\frac{B_{2}A_{2}N_{2}}{k}$

Their ratio of current sensitivity of coil M2 to that of coil M1 is, therefore,  $\frac{B_{2}A_{2}N_{2}}{B_{1}A_{1}N_{1}}$

$=\frac{0.5\times 1.8\times 10^{-3}\times 42 }{0.25\times 3.6\times 10^{-3}\times 30}=1.4$

Determine the ratio of voltage sensitivity of  $M_2$ and $M_1$

The torque experienced by the moving coil M1 for a current I passing through it will be equal to $\tau =B_{1}A_{1}N_{1}I$

The coil will experience a restoring torque proportional to the twist $\phi$

$\phi k=B_{1}A_{1}N_{1}I$

we know V=IR

Therefore, $\phi k=\frac{B_{1}A_{1}N_{1}V}{R_{1}}$

Voltage sensitivity of coil M=$\frac{B_{1}A_{1}N_{1}}{kR_{1}}$

Similarly for coil M2 Voltage sensitivity = $\frac{B_{2}A_{2}N_{2}}{kR_{2}}$

Their ratio of voltage sensitivity of coil M2 to that of coil M1

$=\frac{B_{2}A_{2}N_{2}R_{1}}{B_{1}A_{1}N_{1}R_{2}}$

$=1.4\times \frac{10}{14}$

$\\=1$

The magnetic force on a moving charged particle in a magnetic field is given by $\vec{F_{B}}=q\vec{V}\times \vec{B}$

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)=$6.5 G ( 1G = 10 ^{-4}T )$

Speed of electron(v)=$4.8 \times 10 ^ 6 ms ^{-1}$

Charge of electron=$-1.6\times 10^{-19}C$

Mass of electron=$9.1\times 10^{-31}kg$

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

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

$\\\frac{mV^{2}}{r}=q\vec{V}\times \vec{B}\\ \frac{mV^{2}}{r}=|qVBsin\theta| \\ r=|\frac{mV}{qBsin\theta }|$

$r=\frac{9.1\times 10^{-31}\times 4.8\times 10^{6}}{1.6\times 10^{-19}\times 6.5\times 10^{-4}}=4.2 cm$

In exercise 4.11 we saw $r=\frac{eB}{mv}$

Time taken in covering the circular path once(time period (T))=$\frac{2\pi r}{v} =\frac{2\pi m}{eB}$

Frequency, $\nu =\frac{1}{T}=\frac{eB}{2\pi m}$

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

$\nu =\frac{1.6\times 10^{-19}\times 4.8\times 10^{6}}{2\pi \times 9.1\times 10^{-31}}=18.2 MHz$

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=60o

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.

$\\\tau =nBIAsin\theta\\ \tau = 30\times 1\times 6\times \pi \times (0.08)^{2}\times sin60^{o}$

=3.13 Nm

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

From the relation $\tau =nBIAsin\theta$ 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

$B=\frac{\mu _{o}I}{2r}$

$\\B_{x}=\frac{n_{x}\mu _{o}I_{x}}{2r_{x}}\\ \\B_{x}=\frac{20\times 4\pi \times 10^{-7}\times 16}{2\times 0.16}$

$\\B_x=4\pi\times 10^{-4}T$(towards East)

$\\B_{y}=\frac{n_{y}\mu _{o}I_{y}}{2r_{y}}\\ \\B_{y}=\frac{25\times 4\pi \times 10^{-7}\times 18}{2\times 0.1}$

$\\B_y=9\pi \times10^{-4}T$(towards west)

The net magnetic field at the centre of the coils,

Bnet=B- Bx

=1.57$\times$10-3 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  $100 G ( 1G = 10 ^{-4}) T$

$B=\mu _{o}nI$

$nI=\frac{B}{\mu _{o}}$

nI$\\=\frac{100\times 10^{-4}}{4\pi \times 10^{-7}}\\ \\=7957.74 \approx 8000$

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.

$B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}}$

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,

$B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}}$

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

$B = \frac{\mu _0 IN}{2R}$

$B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}}$

Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,
$B = 0.72 \frac{\mu _0 NI}{R}$  approximately.

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

.$B_{1}= \frac{\mu _0 IR^2N}{2 ( (R/2+l)^2 + R^2 )^{3/2}}$

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

$B_{2}= \frac{\mu _0 IR^2N}{2 ( (R/2-l)^2 + R^2 )^{3/2}}$

Since the direction of current in both the coils is same the magnetic fields B1 and B2 due to them at point P would be in the same direction

Bnet =B1+B2

$\\B_{net}= \frac{\mu _0 IR^2N}{2 ( (R/2-l)^2 + R^2 )^{3/2}}+\frac{\mu _0 IR^2N}{2 ( (R/2+l)^2 + R^2 )^{3/2}}\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( (R/2-l)^2 + R^2 )^{-3/2} + ( (R/2+l)^2 + R^2 )^{-3/2}\right ]\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( \frac{R^{2}}{4}-Rl+l^{2} + R^2 )^{-3/2} + ( \frac{R^{2}}{4}+Rl+l^{2} + R^2 )^{-3/2}\right ]\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( \frac{5R^{2}}{4}-Rl+l^{2} )^{-3/2} + ( \frac{5R^{2}}{4}+Rl+l^{2} )^{-3/2}\right ]\\$

$\\B_{net}= \frac{\mu _0 IR^2N}{2}\times (\frac{5R^{2}}{4})^{-3/2}\left [ ( 1-\frac{4l}{5R}+\frac{4l^{2}}{5R^{2}} )^{-3/2} + ( 1+\frac{4l}{5R}+\frac{4l^{2}}{5R^{2}} )^{-3/2}\right ]\\$

Since l<<R we can ignore term l2/R2

$\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{5}{4})^{-3/2}\left [ ( 1-\frac{4l}{5R} )^{-3/2} + ( 1+\frac{4l}{5R} )^{-3/2}\right ]\\$

$\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{4}{5})^{3/2}\left [ 1+\frac{6l}{5R} + 1-\frac{6l}{5R} \right ]\\$

$\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{4}{5})^{3/2}\times 2$

$\\B_{net}= 0.715\frac{\mu _0 IN}{R}\approx 0.72\frac{\mu _0 IN}{R}$

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

$B=\frac{\mu _{o}NI}{l}$

Total number of turns(N)=3500

Current flowing in toroid =11 A

Length of the toroid, l=

$\\l=2\pi \left (\frac{ r_{1}+r_{2}}{2} \right )\\ \\l=\pi ( r_{1}+r_{2})\\ \\l=\pi (0.25+0.26)\\ \\l=0.51\pi$ (r1=inner radius=25 cm, r2=outer radius=26 cm)

$B=\frac{4\pi \times 10^{-7}\times 3500\times 11}{0.51\pi }=0.031 T$

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 $q\vec{V}$ is towards the west direction, Force will be towards south direction if the magnetic field is in the vertically downward direction as $\vec{F}=q\vec{V}\times \vec{B}$

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

Therefore K.E of electron = $1.6$$\times$$10^{-19}$$\times$$2000$$=3.2$$\times10^{-16} J$

$\\\frac{1}{2}mv^{2}=3.2\times 10^{-16}\\ \\v=\sqrt{\frac{2\times 3.2\times 10^{-16}}{9.1\times 10^{-31}}}\\ v=2.67\times 10^{7}\ ms^{-1}$

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,

$\\\frac{mv^{2}}{r}=evB\\ r=\frac{mv}{eB}\\ r=\frac{9.1\times 10^{-31}\times 2.67\times 10^{7}}{1.6\times 10^{-19}\times 0.15}=1.01mm$

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

Therefore K.E of electron = 1.6$\times$10-19$\times$2000=3.2$\times$10-16 J

$\\\frac{1}{2}mv^{2}=3.2\times 10^{-16}\\ \\v=\sqrt{\frac{2\times 3.2\times 10^{-16}}{9.1\times 10^{-31}}}\\ v=2.67\times 10^{7}\ ms^{-1}$

The component of velocity perpendicular to the magnetic field is

$\\v_{p}=vsin30^{o}\\ v_{p}=1.33\times 10^{7}\ ms^{-1}$

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

$\\\frac{mv^{2}_{p}}{r}=ev_{p}B\\ r=\frac{mv_{p}}{eB}\\ r=\frac{9.1\times 10^{-31}\times 1.33\times 10^{7}}{1.6\times 10^{-19}\times 0.15}$

r=5m

r=5$\times$10-4 m

r=0.5 mm

The component of velocity along the magnetic field is

$\\v_{t}=vcos30^{o}\\ v_{t}=2.31\times 10^{7}\ ms^{-1}$

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

$\\p=\frac{2\pi r}{v_{p}}\times v_{t}\\ p=\frac{2\pi \times 5\times 10^{-4}}{1.33\times 10^{7}}\times 2.31\times 10^{7}$

p=5.45$\times$10-3 m

p=5.45 mm

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

$\\qE=qvB\\ E=vB$ (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,

$\\\frac{1}{2}mv^{2}=qV\\ \\v=\sqrt{\frac{2qV}{m}}$(ii)

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

$\\E=B\sqrt{\frac{2qV}{m}}\\ \frac{q}{m}=\frac{E^{2}}{2VB^{2}}\\ \frac{q}{m}=\frac{(9\times 10^{-5})^{2}}{2\times 15\times 10^{3}\times (0.75)^{2}}=4.8\times 10^{-13}$

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.

$mg=BIl$

mass of rod=0.06 g

length of rod=0.45m

the current flowing through the rod=5 A

$\\B=\frac{mg}{Il}\\ B=\frac{0.06\times 9.8}{5\times 0.45}\\ B=0.261\ T$

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

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

$\\T=mg+BIl\\ \\T=0.06\times 9.8+0.261\times 5\times 0.45\\ \\T=1.176\ N$

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.

$F=\frac{\mu _{o}I_{1}I_{2}}{2\pi d}$

Current in both wires=300 A

Distance between the wires=1.5 cm

Permeability of free space=4$\pi \times$10-7 TmA-1

$F=\frac{4\pi \times 10^{-7}\times 300\times 300}{2\pi \times 0.015}$

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=90o

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

$\\F=BIlsin\theta \\ F=1.5\times 7\times 0.2$

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=45o

The radius of the cylindrical region=10.0 cm

The length of wire inside the magnetic field, $\\l=\frac{2r}{sin\theta }\\$

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

$\\F=BIlsin\theta \\ \\F=1.5\times 7\times \frac{2\times 0.1}{sin45^{o}}\times sin45^{o}$

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.

$\\(\frac{l}{2})^{2}+d^{2}=r^{2}\\ \\(\frac{l}{2})^{2}=0.1^{2}-0.06^{2}\\ \\(\frac{l}{2})^{2}=0.01-0.0036\\ \\\\(\frac{l}{2})^{2}=0.0064\\ \\\frac{l}{2}=0.08\\ \\l=0.16m$

$\\F=BIlsin\theta \\ F=1.5\times 7\times 0.16$

F=1.68 N

This force acts in the vertically downward direction.

The magnetic field is

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

Area of the loop = length$\times$breadth

A=0.1$\times$0.05

A=0.005 m2

$\vec{A}=0.005\ m^{2}\hat{i}$

$\\\vec{\tau }=I\vec{A}\times \vec{B}\\ \\\vec{\tau }=12\times 0.005\hat{i}\times 0.3\hat{k}\\ \\\vec{\tau }=-0.018\hat{j}$

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

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

Area of the loop = length$\times$breadth

A=0.1$\times$0.05

A=0.005 m2

$\vec{A}=0.005\ m^{2}\hat{i}$ (same as that in the last case)

$\\\vec{\tau }=I\vec{A}\times \vec{B}\\ \\\vec{\tau }=12\times 0.005\hat{i}\times 0.3\hat{k}\\ \\\vec{\tau }=-0.018\hat{j}$

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

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

Area of the loop = length$\times$breadth

A=0.1$\times$0.05

A=0.005 m2

$\vec{A}=-0.005\ m^{2}\hat{j}$

$\\\vec{\tau }=I\vec{A}\times \vec{B}\\ \\\vec{\tau }=12\times- 0.005\hat{j}\times 0.3\hat{k}\\ \\\vec{\tau }=-0.018\hat{i}$

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

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

Area of the loop = length$\times$breadth

A=0.1$\times$0.05

A=0.005 m2

$\vec{A}=0.005\ m^{2}(-\frac{\hat{i}}{2}+\frac{\sqrt{3}}{2}\hat{j})$

$\\\vec{\tau }=I\vec{A}\times \vec{B}\\ \\\vec{\tau }=12\times 0.005(-\frac{\hat{i}}{2}+\frac{\sqrt{3}}{2}\hat{j})\times 0.3\hat{k}\\ \\\vec{\tau }=-0.018(\frac{\hat{i}}{2}+\frac{\sqrt{3}}{2}\hat{j})$

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

The magnetic field is

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

Area of the loop = length$\times$breadth

A=0.1$\times$0.05

A=0.005 m2

$\vec{A}=0.005\ m^{2}\hat{k}$

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

$\vec{B}=3000\ G \hat{k}=0.3\ T\hat{k}$

Current in the loop=12 A

Area of the loop = length$\times$breadth

A=0.1$\times$0.05

A=0.005 m2

$\vec{A}=-0.005\ m^{2}\hat{k}$

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

$\\\vec{\tau }=I\vec{A}\times \vec{B}\\$

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 eVdB where Vd is the drift velocity of the electrons.

The current is given by

$I=neAV_{d}$

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

$\\V_{d}=\frac{I}{neA}\\ \\V_{d}=\frac{5}{10^{29}\times 1.6\times 10^{-19}\times 10^{-5}}$

$Vd=3.125\times10^{-5} ms^{-1}$

The average force on each electron is

$F=eV_dB$

$F=1.6\times10^{-19}\times3.125\times10^{-5}\times00.1$

$F=5\times10^{-25} N$

The magnetic field inside the solenoid is given by

$B=\mu _{0}nI$

n is number of turns per unit length

$n=\frac{3\times 300}{0.6}$

n=1500 m-1

Current in the wire I= 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

$BI_wl=mg$

$\\B=\frac{mg}{I_{w}l}\\ \\\mu _{0}nI=\frac{mg}{I_{w}l}\\ I=\frac{mg}{I_{w}l\mu _{0}n}\\ I=\frac{2.5\times 10^{-3}\times 9.8}{6\times 0.02\times 4\pi \times 10^{-7}\times 1500}\\ I=108.37\ A$

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 =  $12 \Omega$

$I$$\times$$(R+G)=18 V$

$R=\frac{18}{3\times 10^{-3}}-12$

$R=6000-12$

$\\=5988\Omega$

The galvanometer can be converted into a voltmeter by connecting a resistor of resistance $5988\Omega$ 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 =  $15 \Omega$

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

$IG=(6-I)R$

$\\R=\frac{IG}{6-I}\\ \\R=\frac{4\times 10^{-3}\times 15}{6-4\times 10^{-3}}\\ R\approx 0.01\Omega$

The galvanometer can be converted into an ammeter by connecting a resistor of resistance $0.01\Omega$ in parallel with it.

NCERT solutions for class 12 physics chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## Importance of NCERT solutions for class 12 physics chapter 4 moving charges and magnetism in exams:

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.