Magnetic Effects Of Current Question 435
Question: A coaxial cable consists of a thin inner conductor fixed along the axis of a hollow outer conductor. The two conductors carry equal currents in opposites directions. Let $ B_{1} $ and $ B_{2} $ be the magnetic fields in the region between the conductors and outside the conductor, respectively Then,
Options:
A) $ B_{1}\ne 0,,B_{2}\ne 0 $
B) $ B_{1}=B_{2}=0 $
C) $ B_{1}\ne 0,B_{2}=0 $
D) $ B_{1}=0,B_{2}\ne 0 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Apply Ampere’s circular law to the coaxial circular loops $ L_{1} $ and $ L_{2} $ . The magnetic field is $ B_{1} $ at all points on $ L_{1} $ and $ B_{2} $ at all points on $ L_{2} $ . $ \sum I\ne 0 $ for $ L_{1} $ and 0 for $ L_{2} $ Hence, $ B_{1}\ne 0 $ but $ B_{2}=0 $ $ [ As,\oint{\overset{\to }{\mathop{B}},.d,\overset{\to }{\mathop{i}},={\mu_{0}}\sum ,I} ] $
BETA
