Class 12 Physics Chapter 4 - Moving Charges and Magnetism NCERT Solutions: Questions and Answers PDF

4.1 Number of turns of the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given as

$\left|\stackrel{?}{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2\pi nI}{r}$ where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$
Hence, $\left|\stackrel{?}{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\pi \times 100\times 0.4}{0.08}$ = 3.14 $\times {10}^{-4}$ T

4.2 Current in the wire, I = 35 A

Distance of a point from the wire, r = 20 cm = 0.2 m

Magnitude of the magnetic field at this point is given as

$\left|\stackrel{?}{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$ where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

Hence, $\left|\stackrel{?}{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\times 35}{0.2}$ = 3.50 $\times {10}^{-5}$ T

4.3 Current in the wire, I = 50 A

The distance of the point from the wire, r = 2.5 m

Magnitude of the magnetic field at this point is given as

$\left|\stackrel{?}{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$ where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

Hence, $\left|\stackrel{?}{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\times 50}{2.5}$ = 4.0 $\times {10}^{-6}$ T

The direction of the current in the wire is vertically downward. Hence, according to Maxwell’s right hand rule, the direction of the magnetic field at the given point is vertically upward.

4.4 Current in the power line, I = 90 A

Point is located below power line at a distance, r = 1.5 m

Magnitude of the magnetic field at this point is given as

$\left|\stackrel{?}{B}\right|$ = $\frac{{\mu }_0}{4\pi }\frac{2I}{r}$ where ${\mu }_0$ = Permeability of free space $\pi \times {10}^{-7}$ = 4 Tm $A^{-1}$

Hence, $\left|\stackrel{?}{B}\right|$ = $\frac{4\pi \times {10}^{-7}\mathrm{}}{4\pi }\frac{2\times 90}{1.5}$ = 1.2 $\times {10}^{-5}$ T

4.5 Current in the wire, I = 8 A

Magnitude of the uniform magnetic field, B = 0.15 T

Angle between the wire and the magnetic field, $\theta$ = 30 $°$

Magnetic force per unit length of the wire is given as

f = BI $\sin ?\theta$ = 0.15 $\times 8\times \sin ?30°$ = 0.6 N/m

4.6 Length of the wire, l = 3 cm = 0.03 m

Current flowing in the wire, I = 10 A

Magnetic field, B = 0.27 T

Angle between current and the magnetic field, $\theta$ = 90 $°$

Magnetic force exerted on the wire is given as

F= BI $l\sin ?\theta$ = 0.27 $\times 10\times 0.03\sin ?90°$ = 8.1 $\times {10}^{-2}$ N

4.7 Current flowing in wire A, $I_A$ = 8.0 A

Current flowing in wire B, $I_B$ = 5.0 A

Distance between two wires, r = 4.0 cm = 0.04 m

Length of the section of the wire A, L = 10 cm = 0.1 m

Force exerted on length L due to magnetic field is given as:

F = $\frac{{\mu }_0{I}_A{I}_{B}L}{2\pi r}$ , where ${\mu }_0$ = permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

F = $\frac{4\pi \times {10}^{-7}\times 8\times 5\times 0.1}{2\pi \times 0.04}$ = 2 $\times {10}^{-5}$ N

The magnitude of force is 2 $\times {10}^{-5}$ N. This is an attractive force normal to A, towards B. Because the direction of the currents in both the wire is same.

4.8 Length of the solenoid, l = 80 cm = 0.8 m

Total number of turns in 5 layers, n = 5 $\times 400$ = 2000

Diameter of the solenoid, D = 1.8 cm = 0.018 m

Current carrying by the solenoid, I = 8.0 A

Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

B = $\frac{{\mu }_{0}NI}{l}$ , where ${\mu }_0$ = permeability of free space = 4 $\pi \times {10}^{-7}$ Tm $A^{-1}$

B = $\frac{4\pi \times {10}^{-7}\times 2000\times 8}{0.8}$ = 2.51 $\times {10}^{-2}$ T

4.9 Length of a side of the square coil, l = 10 cm = 0.1 m

Current flowing through the coil, I = 12 A

Number of turns of the coil, n = 20

Angle made by the plane of the coil with magnetic field, $\theta$ = 30 $°$

Strength of the magnetic field, B = 0.80 T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by,

$\tau$ = nBIA $\sin ?\theta$ , where A = Area of the square coil = 0.1 $\times 0.1=0.01m^2$

$\tau =20\times 0.8\times 12\times 0.01\times \sin ?30°$ = 0.96 Nm

4.10 For moving coil meter M1

Current sensitivity of $M_1$ is given as $I_{s1}$ = $\frac{N_1{B}_1{A}_1}{K_1}$ and

for $M_2$ is given as $I_{s2}$ = $\frac{N_2{B}_2{A}_2}{K_2}$ where $K_1$ and $K_2$ are spring constants for both $M_1$ & $M_2$ . It is given $K_1$ = $K_2$

The ratio of current sensitivity is given as $\frac{I_{s2}\mathrm{}}{I_{s1}}$ = $\frac{N_2{B}_2{A}_2}{N_1{B}_1{A}_1}$ = $\frac{42\times 0.5\times 1.8\mathrm{}\times \mathrm{}{10}^{-3}}{30\times 0.25\times 3.6\times {10}^{-3}}$ = 1.4

Voltage sensitivity of $M_{1}and{M}_2$ is given by $V_{S1}$ = $\frac{N_1{B}_1{A}_1}{K_1{R}_1}$ and $V_{S2}$ = $\frac{N_2{B}_2{A}_2}{K_2{R}_2}$

The ratio of voltage sensitivity $\frac{V_{S2}}{V_{S1}}$ = $\frac{N_2{B}_2{A}_2{K}_1{R}_1}{N_1{B}_1{A}_1{K}_2{R}_2}$ = $\frac{N_2{B}_2{A}_2{R}_1}{N_1{B}_1{A}_1{R}_2}$

= $\frac{42\times 0.5\times 1.8\mathrm{}\times \mathrm{}{10}^{-3}\times 10}{30\times 0.25\times 3.6\times {10}^{-3}\times 14}$ = 1

4.11 Magnetic field strength, B = 6.5 G = 6.5 $\times {10}^{-4}$ T

Speed of electron, v = 4.8 $\times {10}^6$ m/s

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

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

Angle between the shot electron and the magnetic field, $\theta$ = 90 $°$

Magnetic force exerted on the electron in the magnetic field is given as:

F = evB $\sin ?\theta$

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r

The centripetal force exerted on electron, $F_c$ = $\frac{m{v}^2}{r}$

In equilibrium, the centripetal force exerted on electron = magnetic force on the electron

F = $F_c$

evB $\sin ?\theta =\frac{m{v}^2}{r}$

r = $\frac{mv}{\mathrm{e}\mathrm{B}\sin ?\theta }$ = $\frac{9.1\mathrm{}\times {10}^{-31}\times 4.8\mathrm{}\times {10}^6}{1.6\mathrm{}\times {10}^{-19}\times 6.5\times {10}^{-4}\times sin90°}$ = 4.20 $\times {10}^{-2}$ m = 4.20 cm

4.12 Let the frequency of revolution = $\nu$

Angular frequency = $\omega =2\pi \nu$

Now, velocity of electron, v = r $\omega$

Since in circular orbit, magnetic force is balanced by the centripetal force, we can write

evB $=\frac{m{v}^2}{r}$ or eB = $\frac{mv}{r}$ = $\frac{m\left(\mathrm{r}\omega \right)}{r}$ = $\frac{2\pi \nu mr}{r}$

This frequency is independent of the speed of electron.

$\nu =\frac{1.6\mathrm{}\times {10}^{-19}\times 6.5\times {10}^{-4}}{2\pi \times 9.1\mathrm{}\times {10}^{-31}}=18.19\times {10}^6$ Hz = 18.19 MHz

4.13 Number of turns of the circular coil, n = 30

Radius of the coil, r = 8.0 cm = 0.08 m

Area of the coil, A = $\pi r^2$ = $\pi \times {0.08}^2$ = 0.0201 $m^2$

Current flowing in the coil, I = 6.0 A

Magnetic field strength, B = 1 T

Angle between the field line and normal of the coil surface, $\theta$ = 60°

The coil experiences a toque in the magnetic field, hence it turns.

The counter torque is given by the relation,

$\tau =nIBA\sin ?\theta$ = 30 $\times 6\times 1\times 0.0201\sin ?60°$ = 3.133 N

Since the magnitude of the torque is not dependent on the shape of the coil, it depends only on the area. Hence he answer will not change if the circular coil is replaced with a planar coil of same area.

4.14 For coil X:

Radius, $r_1$ = 16 cm = 0.16 m

Number of turns, $n_1$ = 20

Current, $I_1$ = 16 A

For coil Y:

Radius, $r_2$ = 10 cm = 0.10 m

Number of turns, $n_2$ = 25

Current, $I_2$ = 18 A

Magnetic field due to coil X at their centre is given by the relation:

$B_1$ = $\frac{{\mu }_0{n}_1{I}_1}{2r_1}$ , where = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$
$B_1=\frac{4\pi \times {10}^{-7}\mathrm{}\times 20\times 16}{2\times 0.16}$ = 1.257 $\times {10}^{-3}$ T (towards East)

Magnetic field due to coil Y at their centre is given by the relation:

$B_2$ = $\frac{{\mu }_0{n}_2{I}_2}{2r_2}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

$B_2=\frac{4\pi \times {10}^{-7}\mathrm{}\times 25\times 18}{2\times 0.10}$ = 2.827 $\times {10}^{-3}$ T (towards West)

The net magnetic field B = $B_2$ - $B_1$ = 2.827 $\times {10}^{-3}$ - 1.257 $\times {10}^{-3}$ = 1.57 $\times {10}^{-3}$ T ( towards West)

4.15 Magnetic field strength, B = 100 G = 100 ${10}^{-4}$ T

Number of turns per unit length, n = 1000 turns / m

Current flowing in the coil, I = 15 A

${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Magnetic field is given by the relation,

B = ${\mu }_{0}nI$ or nI = $\frac{B}{{\mu }_0}$ = $\frac{100\times \mathrm{}{10}^{-4}}{4\pi \times {10}^{-7}}$ = 7957.75 A $m^{-1}\approx 8000$ A $m^{-1}$

If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400 and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.

4.16 Radius of the circular coil = R

Number of turns on the coil = N

Current in the coil= I

Magnetic field at a point on its axis at a distance x is given as:

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

where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

If the magnetic field at the centre of the coil is considered, then x = 0, then

B = $\frac{{\mathit{\mu }}_0\mathit{I}{\mathit{R}}^2\mathit{N}}{2({0+\mathit{}{\mathit{R}}^2)}^{\frac{3}{2}}}$ = $\frac{{\mathit{\mu }}_0\mathit{I}\mathit{N}}{2\mathit{R}}$

This is the familiar result for magnetic field at the centre of the coil.

Radius of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both the coils = I $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{{\left\{\right(\frac{R}{2}-d)}^2+R^2\}}^{\frac{3}{2}}+{{\left\{\right(\frac{R}{2}+d)}^2+R^2\}}^{\frac{3}{2}}\right]$

Distance between both the coils = R

Let us consider point Q at a distance d from the centre.

Then one coil is at a distance of + d from point Q

Magnetic field at point Q is given as:

$B_1$ = $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$

$\mathrm{}\mathrm{A}\mathrm{l}\mathrm{s}\mathrm{o}$ , the other coil is at a distance of $\frac{R}{2}$ – d from point Q,

Magnetic field due to this coil is given as

$B_2$ = $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$

Total magnetic field, B = $B_1$ + $B_2$

= $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}+d)}^2+R^2)}^{\frac{3}{2}}}$ + $\frac{{\mu }_{0}I{R}^{2}N}{2({{\frac{R}{2}-d)}^2+R^2)}^{\frac{3}{2}}}$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{{\left\{\right(\frac{R}{2}-d)}^2+R^2\}}^{\frac{3}{2}}+{{\left\{\right(\frac{R}{2}+d)}^2+R^2\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\left[{\{\frac{5R^2}{4}+d^2-Rd\}}^{\frac{3}{2}}+{\{\frac{5R^2}{4}+d^2+Rd\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{5R^2}{4}\right)}^{\frac{3}{2}}\left[{\{1+\frac{4}{5}\frac{d^2}{R^2}-\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}+{\{1+\frac{4}{5}\frac{d^2}{R^2}+\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}\right]$

For d $?R$ , neglecting the factor , we get

= $\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{5R^2}{4}\right)}^{\frac{3}{2}}\left[{\{1-\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}+{\{1+\frac{4}{5}\frac{d}{R}\}}^{\frac{3}{2}}\right]$

= $\frac{{\mu }_{0}I{R}^{2}N}{2R^3}\times {\left(\frac{4}{5}\right)}^{\frac{3}{2}}\left[1-\frac{6d}{5R}+1+\frac{6d}{5R}\right]$

= ${\left(\frac{4}{5}\right)}^{\frac{3}{2}}\frac{{\mu }_{0}IN}{R}$

B = 0.72 $\frac{{\mu }_{0}IN}{R}$

Hence, it is proved that the field on the axis around mid-point between the coils is uniform.

4.17 Inner radius of the toroid, $r_1$ = 25 cm = 0.25 m

Outer radius of the toroid, $r_2$ = 26 cm = 0.26 m

Number of turns on the coil, N = 3500

Current in the coil, I = 11 A

Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.

Magnetic field inside the core of a toroid is given by the relation,

B = $\frac{{\mu }_{0}NI}{l}$ . where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

L = length of the toroid = 2 $\pi (\frac{r_1+r_2}{2}$ ) = $\pi$ (0.25 + 0.26)= 1.6022

B = $\left(\frac{4\pi \times {10}^{-7}\times 3500\times 11}{1.6022}\right)$ = 3.0 $\times {10}^{-2}$ T

Magnetic field in the empty space surrounded by the toroid is zero.

4.18 (a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.

(b) Yes, the final speed of the particle will be equal to its initial speed. This because magnetic force can change the direction of velocity, not its magnitude.

(c) This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the south. According to Fleming's left hand rule, magnetic field should be applied in a vertically downward direction.

4.19 Magnetic field strength, B = 0.15 T

Charge on the electron, e = 1.6 $\times {10}^{-19}$ C

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

Potential difference, V = 2.0 kV = 2 $\times {10}^3$ V

Thus the kinetic energy of the electron = eV = $\frac{1}{2}$ m $v^2$ , where v = velocity of electron

v = $\sqrt{\frac{2eV}{m}}$ ……..(1)

Magnetic force on the electron provides the required centripetal force of the electron. Hence, electron traces a circular path of radius r

Magnetic force on the electron = Bev

Centripetal force $\frac{m{v}^2}{r}$

Hence, Bev = $\frac{m{v}^2}{r}$

r = $\frac{mv}{Be}$ ………………(2)

From equation (1) and (2), we get

r = $\frac{m}{Be}{\left[\frac{2eV}{m}\right]}^{\frac{1}{2}}$

= $\frac{9.1\mathrm{}\times {10}^{-31}}{0.15\times 1.6\mathrm{}\times {10}^{-19}}\times$ ${\left[\frac{2\times 1.6\mathrm{}\times {10}^{-19}\times 2\mathrm{}\times {10}^3}{9.1\mathrm{}\times {10}^{-31}}\right]}^{\frac{1}{2}}$

r = 1.006 $\times {10}^{-3}$ m = 1.0 mm

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

When the field makes an angle $\theta$ of 30 $°$ wit initial velocity, the initial velocity will be,

$v_1$ = v $\sin ?\theta$

$\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{m}$ equation (2), we can write the expression for the new radius as :

$r_1=\frac{m{v}_1}{Be}=\frac{mv\sin ?\theta }{Be}=$ $\frac{9.1\mathrm{}\times {10}^{-31}}{0.15\times 1.6\mathrm{}\times {10}^{-19}}\times$ ${\left[\frac{2\times 1.6\mathrm{}\times {10}^{-19}\times 2\mathrm{}\times {10}^3}{9.1\mathrm{}\times {10}^{-31}}\right]}^{\frac{1}{2}}\times \sin ?30°$

Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.

4.20 Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 $\times {10}^{-3}$ V

Electrostatic field, E = 9.0 $\times {10}^5$ V/m

Let the mass of electron = m, Charge of the electron = e, Velocity of the electron = v

Then kinetic energy of the electron = eV

$\frac{1}{2}$ m $v^2$ = eV or $\frac{e}{m}$ = $\frac{v^2}{2V}$ ………….(1)

4.21 Length of the rod. l= 0.45 m

Mass suspended by the wire, m = 60 g = 60 $\times {10}^{-3}$ kg

Acceleration due to gravity, g = 9.8 m/ $s^2$

Current, I = 5 A

To achieve zero tension, the magnetic field = weight of the wire

BIl = mg or

B = $\frac{mg}{Il}$ = $\frac{60\mathrm{}\times {10}^{-3}\times 9.8}{5\times 0.45}$ = 0.26 T

The magnetic field should be set up such that it gives an upward magnetic force.

If the direction of current is reversed, then the magnetic force will act downwards and total tension in the wire will be

mg + BIl = 60 $\times {10}^{-3}\times 9.8+0.26\times 5\times 0.45$ = 1.173 T

4.22 Current in both the wires, I = 300 A

Distance between the wires, r = 1.5 cm = 0.015 m

Length of the two wires, l = 70 cm = 0.7 m

Now, force between the two wires is given by the relation:

F = $\frac{{\mu }_0{I}^2}{2\pi r}$ , where ${\mu }_0$ = Permeability of free space = 4 $\pi \times {10}^{-7}$ T m $A^{-1}$

Hence F = $\frac{4\pi \times {10}^{-7}\times {300}^2}{2\pi \times 0.015}$ N/m = 1.2 N/m

4.24 Magnetic field strength, B = 3000G = 3000 $\times {10}^{-4}$ T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop, A = l $\times b$ = 10 $\times 5$ = 50 ${cm}^2$ = 50 $\times {10}^{-4}$ $m^2$

Current in the loop, I = 12 A

Assume that the anti-clockwise direction of the current is positive and vice versa.

Torque, $\stackrel{?}{\tau }$ = $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

From the given figure, it can be observed that A is normal to the y-z plane and B is directed along z-axis.

$\tau$ = 12*( 50 $\times {10}^{-4})\stackrel{?}{i}\times 0.3\stackrel{?}{k}$ = $-1.8\times {10}^{-2}\stackrel{?}{j}$ Nm

The Torque $1.8\times {10}^{-2}$ is Nm along the negative y-direction.

The force on the loop is zero because the angle between A & B is zero.

This case is similar to case (a). The answer is same as case (a)

Torque, $\stackrel{?}{\tau }$ = I $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

From the given figure, it can be observed that A is normal to the x-z plane and B is directed along z-axis.

= - 12 $\times$ (50 $\times {10}^{-4})\stackrel{?}{i}\times 0.3\stackrel{?}{k}$ = $1.8\times {10}^{-2}$ Nm

The Torque is $1.8\times {10}^{-2}$ Nm at an angle 240 $°$ with positive x direction. The force is zero

Magnitude of the torque is given as:

| $\left|\tau \right|$ |= IAB|

12 $\times$ 50 $\times {10}^{-4}\times 0.3$ = $1.8\times {10}^{-2}$ Nm

The Torque is $1.8\times {10}^{-2}$ is Nm at an angle 240 with positive x direction. The force is zero.

Torque, $\left|\tau \right|$ = I $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

= (12 $\times$ 50 $\times {10}^{-4})\stackrel{?}{k}$ $\times 0.3\stackrel{?}{k}$

= 0

The torque is zero and the force is also zero.

Torque, $\stackrel{?}{\tau }$ = I $\stackrel{?}{A}$ $\times \stackrel{?}{B}$

= (12 $\times$ 50 $\times {10}^{-4})\stackrel{?}{k}$ $\times 0.3\stackrel{?}{k}$

= 0

The torque is zero and the force is also zero.

In the case of (e), the direction of I $\stackrel{?}{A}$ and $\stackrel{?}{B}$ is the same and the angle between them is zero. If displaced, they come back to equilibrium. Hence, its equilibrium is stable.

In the case of (f), the direction of I $\stackrel{?}{A}$ and $\stackrel{?}{B}$ is opposite. The angle between them is 180 $°$ . If disturbed, it does not come back to its original position, hence the equilibrium is unstable.

4.25 Number of turns on the circular coil, n =20

Radius of the coil, r = 10 cm = 0.1 m

Magnetic field strength, B = 0.10 T

Current in the coil, I = 5.0 A

The total torque on the coil is zero because the field is uniform.

The total torque on the coil is zero because the field is uniform.

Cross-sectional area of the copper coil, A = ${10}^{-5}$ $m^2$

Number of free electrons per cubic meter of copper, N = ${10}^{29}$ / $m^3$

Charge on the electron, e = 1.6 $\times {10}^{19}$ C

Magnetic force, F = Be $v_d$ where $v_d$ is the drift velocity of electrons

$v_d=\frac{I}{NeA}$ ${\mu }_0$

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}F$ = $\frac{BeI}{NeA}$ = $\frac{0.1\times 5}{{10}^{29}\times {10}^{-5}}$ = 5 $\times {10}^{-25}$ N

The average force on each electron is 5 $\times {10}^{-25}$ N

4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 $\times {10}^{-3}$ kg $\times {10}^{-3}$

Hence, the current flowing through the solenoid is 108 A.

4.27 Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, $I_g$ = 3 mA = 3 $\times {10}^{-3}$ A

Range of voltmeter = 0, to be converted to 18 V, hence V = 18 V

Let there be a resistor R connected in series with the galvanometer to convert it into a voltmeter. R is given as

R = $\frac{V}{\mathrm{}{I}_g}$ - G = $\frac{18}{3\mathrm{}\times {10}^{-3}}$ - 12 = 5988 Ω

Hence the required value of resistor is 5988 Ω

4.28 Resistance of the galvanometer coil, G = 15 Ω

Current for which galvanometer shows full deflection, $I_g$ = 4 mA = 4 $\times {10}^{-3}$ A

Range of ammeter has to be converted from 0 to 6 A, hence I = 6 A

A shunt resistor S is to be connected in parallel with the galvanometer to convert it to an ammeter. The value of S is given as

S = 10 mΩ

Hence, a shunt resistor of 10 mΩ is to be connected to galvanometer to convert it to an ammeter.

The different types of magnetism are - Diamagnetism, Paramagnetism, and Ferromagnetism.

Tesla (T) is the SI unit of magnetic field. Tesla is named after Nikola Tesla.

According to this law, at any point, magnetic field due to a tiny current element is directly proportional to the current element length.

There are four types of power supply - switch-mode power supplies, linear power supplies, linear power supplies, and programmable power supplies

There are two types of electricity - static and current. The current electricity is divided into Alternating Current (AC) and Direct Current (DC). Based on the production methods, electricity is divided into - heat, chemical, pressure, light and magnetism.