Class 12th

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

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 $\pi *{10}^{-7}$ Tm $A^{-1}$
Hence, $\left|\stackrel{?}{B}\right|$ = $\frac{4\pi *{10}^{-7}\mathrm{}}{4\pi }\frac{2\pi *100*0.4}{0.08}$ = 3.14 $*{10}^{-4}$ T

3.23 Internal resistance of the cell = r

Balance point of the cell in open circuit, $l_1$ = 76.3 cm

An external resistance R is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, $l_2$ = 64.8 cm

Current flowing through circuit = I

The relation connecting resistance and emf is,

r = ( $\frac{l_1-l_2}{l_2}$ )R = $\frac{76.3-64.8}{64.8}$ $*9.5=$ 1.69 Ω

Therefore, the internal resistance is 1.69 Ω.