A coaxial cable having radii a, b and c carri | vedclass

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(A) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed path is equal to $\mu_0$ times the total cur

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(A) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed path is equal to $\mu_0$ times the total current $i_{	ext{enclosed}}$ passing through the surface bounded by the path.$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 i_{	ext{enclosed}}$For a point $P$ outside the cable at a distance $r > c$ from the axis,we consider a circular Amperian loop of radius $r$.The total current enclosed by this loop is the sum of the current in the inner conductor $(+i)$ and the current in the outer conductor $(-i)$.$i_{	ext{enclosed}} = i + (-i) = 0$Therefore,the magnetic induction $B$ at point $P$ is:$B(2 \pi r) = \mu_0 (0)$$B = 0$

$A$ coaxial cable having radii $a, b$ and $c$ carries equal and opposite currents of magnitude $i$ in the inner and outer conductors. What is the magnitude of the magnetic induction at point $P$ outside of the cable at a distance $r$ from the axis?

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