Moving Charges and Magnetism

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 i

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 ${\mu }_0$ = permea

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

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

4.5 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

4.4 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

4.3 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

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

4.1 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

Absolutely, if thoroughly practiced and understood, these solutions are more than enough to help one score full marks in this chapter. It comprises all key concepts such as laws, derivations, and applications and covers all questions from the NCERT textbook. The step-by-step solutions help students