Atelephone cable at a place has four long straight horizontal wirescarrying a current of 1.0 A in the same direction east to west. Theearth’s magnetic field at the place is 0.39 G, and the angle ofdip is 35º. The magnetic declination is nearly zero. What arethe resultant magnetic fields at points 4.0 cm below the cable?
Atelephone cable at a place has four long straight horizontal wirescarrying a current of 1.0 A in the same direction east to west. Theearth’s magnetic field at the place is 0.39 G, and the angle ofdip is 35º. The magnetic declination is nearly zero. What arethe resultant magnetic fields at points 4.0 cm below the cable?
Numberof horizontal wires in the telephone cable, n= 4
Currentin each wire, I= 1.0 A
Earth’smagnetic field at a location, H= 0.39 G = 0.39 × 10−4T
Angleof dip at the location, δ= 35°
Angleof declination, θ∼0°
Fora point 4 cm below the cable:
Distance,r = 4 cm =0.04 m
Thehorizontal component of earth’s magnetic field can be writtenas:
Hh= Hcosδ− B
Where,
B= Magnetic field at 4 cm due to current Iin the four wires
= Permeability of free space = 4π× 10−7Tm A−1
=0.2 × 10−4T = 0.2 G
∴ Hh= 0.39 cos 35° − 0.2
=0.39 × 0.819 − 0.2 ≈0.12 G
Thevertical component of earth’s magnetic field is given as:
Hv= Hsinδ
=0.39 sin 35° = 0.22 G
The anglemade by the field with its horizontal component is given as:
Theresultant field at the point is given as:
Fora point 4 cm above the cable:
Horizontalcomponent of earth’s magnetic field:
Hh= Hcosδ+ B
=0.39 cos 35° + 0.2 = 0.52 G
Verticalcomponent of earth’s magnetic field:
Hv= Hsinδ
=0.39 sin 35° = 0.22 G
Angle,θ 
= 22.9°
Andresultant field:
Numberof horizontal wires in the telephone cable, n= 4
Currentin each wire, I= 1.0 A
Earth’smagnetic field at a location, H= 0.39 G = 0.39 × 10−4T
Angleof dip at the location, δ= 35°
Angleof declination, θ∼0°
Fora point 4 cm below the cable:
Distance,r = 4 cm =0.04 m
Thehorizontal component of earth’s magnetic field can be writtenas:
Hh= Hcosδ− B
Where,
B= Magnetic field at 4 cm due to current Iin the four wires
= Permeability of free space = 4π× 10−7Tm A−1
=0.2 × 10−4T = 0.2 G
∴ Hh= 0.39 cos 35° − 0.2
=0.39 × 0.819 − 0.2 ≈0.12 G
Thevertical component of earth’s magnetic field is given as:
Hv= Hsinδ
=0.39 sin 35° = 0.22 G
The anglemade by the field with its horizontal component is given as:
Theresultant field at the point is given as:
Fora point 4 cm above the cable:
Horizontalcomponent of earth’s magnetic field:
Hh= Hcosδ+ B
=0.39 cos 35° + 0.2 = 0.52 G
Verticalcomponent of earth’s magnetic field:
Hv= Hsinδ
=0.39 sin 35° = 0.22 G
Angle,θ = 22.9°
Andresultant field:
