A long conductor of circular cross-section with radius r has current density
J(r) = ρ 0 
= 0 
(for r < R/2) into the plane of paper. There is a point P at distance 'a' from the axis of the conductor [a > R]. Two infinitely long thin conducting wires carrying current I 0 in the same direction are placed at distance a from O perpendicular to OP and parallel to con doctor at either side such that the magnetic field at P is zero. Find the current I 0 in the wires and the direction of current as compared with the direction of current in the conductor.
Text Solution
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Sol. Current through conductor,
I = 
J(r) = 
J(r) dr + 
J(r) dr
= 
· 0 · dr + 
dr
= 0 + 
= 
= 
r 0 R 2 Let us consider a circle with center O and radius OP in a plane perpendicular to the conductor. For all points on the circle, due to symmetry, B is same due to the conductor. Applying Ampere's law
· dl = µ 0 I
⇒ B · 2 π a = µ 0 
π r 0 R
2 ⇒ B ′ = 
As current is into the plane B is downward to P. Now field due to wires A 1 and A 2 must cancel B. That is possible when the current in the wires is out of the plane.
Field due to A 1 , 
= 
Due to A 2 , 
= 
Resultant of B 1 and B 2 ,
= 
+ 
= 2 
· 
= 
opposite to 
.
Net field at P, 
+ 
= 0
– 
µ 0 
= 0
I 0 = 
.
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