Class 12th

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

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

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 $\pi *{10}^{-7}$ T m $A^{-1}$

Magnetic field is given by the relation,

B = ${\mu }_{0}nI$ or nI = $\frac{B}{{\mu }_0}$ = $\frac{100*\mathrm{}{10}^{-4}}{4\pi *{10}^{-7}}$ = 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.

4.14 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 $\pi *{10}^{-7}$ T m $A^{-1}$
$B_1=\frac{4\pi *{10}^{-7}\mathrm{}*20*16}{2*0.16}$ = 1.257 $*{10}^{-3}$ 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 $\pi *{10}^{-7}$ T m $A^{-1}$

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

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

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 = $\pi r^2$ = $\pi *{0.08}^2$ = 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 $*6*1*0.0201\sin ?60°$ = 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.

4.12 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.

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

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

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

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

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

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 $\frac{I_{s2}\mathrm{}}{I_{s1}}$ = $\frac{N_2{B}_2{A}_2}{N_1{B}_1{A}_1}$ = $\frac{42*0.5*1.8\mathrm{}*\mathrm{}{10}^{-3}}{30*0.25*3.6*{10}^{-3}}$ = 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 $\frac{V_{S2}}{V_{S1}}$ = $\frac{N_2{B}_2{A}_2{K}_1{R}_1}{N_1{B}_1{A}_1{K}_2{R}_2}$ = $\frac{N_2{B}_2{A}_2{R}_1}{N_1{B}_1{A}_1{R}_2}$

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

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 $*0.1=0.01m^2$

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

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

Total number of turns in 5 layers, n = 5 $*400$ = 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 $\pi *{10}^{-7}$ Tm $A^{-1}$

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