Physics Ncert Solutions Class 12th

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

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

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

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

4.6 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 $*10*0.03\sin ?90°$ = 8.1 $*{10}^{-2}$ N

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 $*8*\sin ?30°$ = 0.6 N/m

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 $\pi *{10}^{-7}$ = 4 Tm $A^{-1}$

Hence, $\left|\stackrel{?}{B}\right|$ = $\frac{4\pi *{10}^{-7}\mathrm{}}{4\pi }\frac{2*90}{1.5}$ = 1.2 $*{10}^{-5}$ 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 $\pi *{10}^{-7}$ Tm $A^{-1}$

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

4.2 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 $\pi *{10}^{-7}$ Tm $A^{-1}$

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