A current I flows along the length of an infi | vedclass

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

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the magnetic field at any point inside the pipe is zero

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(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{	ext{enclosed}}$ passing through the surface bounded by the loop.$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$For a thin-walled pipe carrying current $I$ along its length,any closed loop drawn inside the pipe encloses zero net current $(I_{	ext{enclosed}} = 0)$.Since the current is distributed only on the surface of the pipe,there is no current flowing through the interior region.Therefore,$\oint \vec{B} \cdot d\vec{l} = 0$,which implies that the magnetic field $\vec{B}$ at any point inside the pipe is zero.

$A$ current $I$ flows along the length of an infinitely long,straight,thin-walled pipe. Then:

Similar Questions

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$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

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