A very long coaxial cable consists of an inner thick wire (radius R) carrying a
ID: 1482680 • Letter: A
Question
A very long coaxial cable consists of an inner thick wire (radius R) carrying a current of 5A, and an outer thick wire encasing the inner wire (starting at R and going out to 2R) and carrying 5A in the opposite direction. What statement is true about how the magnetic field changes as a function of distance from the cable centre, r?
Magnetic field increases going from r=0 to R, then decreases from r=R to 2R becoming 0 at r=2R.
Magnetic field is zero for r=0 to R, then increases linearly as we go from r=R to 2R
Magnetic field increases linearly from the centre of the cable, until we reach r=2R.
Magnetic field is 0 everywhere inside the cable.
A.Magnetic field increases going from r=0 to R, then decreases from r=R to 2R becoming 0 at r=2R.
B.Magnetic field is zero for r=0 to R, then increases linearly as we go from r=R to 2R
C.Magnetic field increases linearly from the centre of the cable, until we reach r=2R.
D.Magnetic field is 0 everywhere inside the cable.
Explanation / Answer
Magnetic field is zero for r=0 to R, then increases linearly as we go from r=R to 2R
From r=0 to r=R:
B = u0 i r/(2 pi a^2)
Therefore the magnetic field is proportional to radius; so "Magnetic field increases going from r=0 to R".
From r=R to r=2R:
B = u0 (i - i (r^2 - a^2)/((2R)^2 - R^2))/(2 pi r)
B = u0 i (1 - (r^2 - a^2)/((2R)^2 - R^2))/(2 pi r)
B = u0 i ({((2R)^2 - a^2) - (r^2 - R^2)}/((2R)^2 - a^2))/(2 pi r)
B = u0 i ((2R)^2 - r^2)/{((2R)^2 - R^2) (2 pi r)}
B = u0 i ((2R)^2)/{((2R)^2 - R^2) (2 pi r)} - u0 i r/{((2R)^2 - R^2) (2 pi)}
B = u0 i (4R^2)/{(3R^2) (2 pi r)} - u0 i r/{(3R^2) (2 pi)}
B = u0 i (4/3)/(2 pi r) - u0 i r/{(3R^2) (2 pi)}
Therefore "decreases from r=R to 2R becoming 0 at r=2R."
B.Magnetic field is zero for r=0 to R, then increases linearly as we go from r=R to 2R
From r=0 to r=R:
B = u0 i r/(2 pi a^2)
Therefore the magnetic field is proportional to radius; so "Magnetic field increases going from r=0 to R".
From r=R to r=2R:
B = u0 (i - i (r^2 - a^2)/((2R)^2 - R^2))/(2 pi r)
B = u0 i (1 - (r^2 - a^2)/((2R)^2 - R^2))/(2 pi r)
B = u0 i ({((2R)^2 - a^2) - (r^2 - R^2)}/((2R)^2 - a^2))/(2 pi r)
B = u0 i ((2R)^2 - r^2)/{((2R)^2 - R^2) (2 pi r)}
B = u0 i ((2R)^2)/{((2R)^2 - R^2) (2 pi r)} - u0 i r/{((2R)^2 - R^2) (2 pi)}
B = u0 i (4R^2)/{(3R^2) (2 pi r)} - u0 i r/{(3R^2) (2 pi)}
B = u0 i (4/3)/(2 pi r) - u0 i r/{(3R^2) (2 pi)}
Therefore "decreases from r=R to 2R becoming 0 at r=2R."