A current i flows in an infinitely long,strai | 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 any point inside a thin-walled pipe,we can choose an Amperian loop (a circle) that lies entirely inside the pipe.Since the current $i$ flows only through the walls of the pipe,the current enclosed by this loop is $i_{	ext{enclosed}} = 0$.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 in an infinitely long,straight and thin-walled pipe. Then,

Similar Questions

An infinitely long hollow conducting cylinder with inner radius $R/2$ and outer radius $R$ carries a uniform current density $J$ along its length. The magnitude of the magnetic field,$|\vec{B}|$ as a function of the radial distance $r$ from the axis is best represented by:

$A$ toroidal solenoid with air core has an average radius $R$,number of turns $N$,and area of cross-section $A$. The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid).

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.

To manufacture a solenoid of length $\ell$ and inductance $L$,the length of the thin wire required is (Diameter of the solenoid is very small compared to its length,$\mu_0$ is the permeability of free space).

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