Ncert Exercises & Solutions

5.19 A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Number of horizontal wires in the telephone cable, n =4
Current in each wire, I= 1.0 A
Earth's magnetic field at a location,

$H = 0.39 G = 0.39 \times 10^{- 4} T$
Angle of dip at the location, õ = 35°
Angle of declination,

$θ$ -0°
Fora point 4 cm below the cable
Distance, r = 4c m, 0.04 m
The horizontal component of the earth's magnetic field can be written as:

$H_{h}$ =H cosõ-B
Where,

$B = Magnetic field at 4 cm due ot current I in the four wires = 4 \times \frac{μ_{0} I}{2 πr}$

$μ_{0} = Permeability of free = 4 π \times 10^{- 7} Tm A^{- 1}$

$\begin{array}{l} ∴ B = 4 \times \frac{4 π \times 10^{- 7} \times 1}{2 π \times 0.04} \\ = 0.2 \times 10^{- 4} T = 0.2 G \\ ∴ H_{h} = 0.39 \cos 35^{\circ} - 0.2 = 0.39 \times 0.819 - 0.2 \approx 0.12 G \end{array}$

$The vertical component of earth' s magnetic field is given as :$

$H_{v} = H \sin δ = 0.39 \sin 35 ° = 0.22 G$

The angle made by the field with its horizontal component is given as

$\begin{aligned} θ & = \tan^{- 1} \frac{H_{v}}{H_{h}} \\ = \tan^{- 1} \frac{0.22}{0.12} = {61.39}^{\circ} \end{aligned}$

The resultant field at the point is given as

$\begin{aligned} H_{1} & = \sqrt{{(H_{v})}^{2} + {(H_{h})}^{2}} \\ = \sqrt{(0.22)^{2} + (0.12)^{2}} = 0.25 G \end{aligned}$

For a point 4 cm above the cable
Horizontal component of earth's magnetic field:

$H_{h} = H \cos δ + B = 0.39 \cos 35 ° + 0.2 = 0.52 G$

Vertical component of earth's magnetic field:

$H_{v} = H \sin δ = 0.39 \sin 35 ° = 0.22 G$

$Angle, θ = \tan^{- 1} \frac{H_{v}}{H_{h}} = \tan^{- 1} \frac{0.22}{0.52} = 22.9 °$

And a resultant field:

$\begin{aligned} H_{2} & = \sqrt{{(H_{v})}^{2} + {(H_{h})}^{2}} \\ = \sqrt{(0.22)^{2} + (0.52)^{2}} = 0.56 T \end{aligned}$

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