First slide

Ampere's circuital law

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 opposite directions. Let B₁ and B₂ be the magnetic fields in the region between the conductors, and outside the conductor, respectively. Then

Moderate

A

$B_{1} \neq B_{2} \neq 0$
B

$B_{1} = B_{2} = 0$
C

$B_{1} \neq B_{2} = 0$
D

$B_{1} = B_{2} \neq 0$

Solution

Apply ampere's circuital law to the coaxial circular loops L₁ and L_2. The magnetic field is B₁ at all points on L₁ and B₂ at all points on $L_{2} . \sum I \neq 0 for L_{1}$ and 0 for L₂.
$\begin{array}{l} Hence B_{1} \neq 0 but B_{2} = 0 \\ [As \oint \vec{B} . d \vec{l} = μ_{0} \sum I] \end{array}$

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