A telephone cable at a place has four long straight horizontal wirescarrying 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 is35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Given: The telephone cable has four wires, current in each wire is 1.0 A in east to west direction, earth’s magnetic field at the place is 0 .39 G and the angle of dip is 35 º .
The magnetic field 4 cm below the cable is given as,
B = n μ 0 I 2 π r
Where, n is the number of horizontal wires in the telephone cable, I is the current in each wire and r is the distance from the cable.
By substituting the given values in the above expression, we get
B = 4 × 4 π × 10 − 7 × 1 2 π × 0.04 = 0.2 × 10 − 4 = 0.2 G
The horizontal component of earth’s magnetic field is given as,
H x = H cos δ − B
Where, H is the earth’s magnetic field strength at the given location and δ is the angle of dip.
By substituting the given values in the above expression, we get
H x = 0.39 cos 35 − 0.2 = 0.12 G
The vertical component of earth’s magnetic field is given as,
H y = H sin δ
By substituting the given values in the above expression, we get
H y = 0.39 sin 35 = 0.22 G
The angle made by the field with its horizontal component is given as,
θ = tan − 1 H y H x
By substituting the given values in the above expression, we get
θ = tan − 1 0.22 0.12 = 62 °
The magnitude of the resultant field is given as,
H 1 = ( H x ) 2 + ( H y ) 2 = ( 0.12 ) 2 + ( 0.22 ) 2 = 0.25 G
Thus, the resultant magnetic field at points 4 .0 cm below the cable is 0 .25 G , 62 ° from the horizontal axis.
For a point 4 .0 cm above the cable, the horizontal component of earth’s magnetic field is given as,
H x = H cos δ + B
By substituting the given values in the above expression, we get
H x = 0.39 cos 35 + 0.2 = 0.52 G
The vertical component of earth’s magnetic field is given as,
H y = H sin δ
By substituting the given values in the above expression, we get
H y = 0.39 sin 35 = 0.22 G
The angle made by the field with its horizontal component is given as,
θ = tan − 1 H y H x
By substituting the given values in the above expression, we get
θ = tan − 1 0.22 0.52 = 23 °
The magnitude of the resultant field is given as,
H 1 = ( H x ) 2 + ( H y ) 2 = ( 0.12 ) 2 + ( 0.52 ) 2 = 0.57 G
Thus, the resultant magnetic field at points 4 .0 cm above the cable is 0 .57 G , 23 ° from the horizontal axis.
Given: The telephone cable has four wires, current in each wire is
The magnetic field
Where,
By substituting the given values in the above expression, we get
The horizontal component of earth’s magnetic field is given as,
Where,
By substituting the given values in the above expression, we get
The vertical component of earth’s magnetic field is given as,
By substituting the given values in the above expression, we get
The angle made by the field with its horizontal component is given as,
By substituting the given values in the above expression, we get
The magnitude of the resultant field is given as,
Thus, the resultant magnetic field at points
For a point
By substituting the given values in the above expression, we get
The vertical component of earth’s magnetic field is given as,
By substituting the given values in the above expression, we get
The angle made by the field with its horizontal component is given as,
By substituting the given values in the above expression, we get
The magnitude of the resultant field is given as,
Thus, the resultant magnetic field at points
