First slide

Ampere's circuital law

Question

In the diagram shown, a wire carries current I. The loop has N turns and part of helical loop on which arrows are drawn is outside the plane of paper. What is the value of the $\oint \vec{B} . \vec{d l}$ ( as in Ampere's law) on the helical loop shown in the figure?

Moderate

A

$- μ_{0} (N I)$
B

$μ_{0} (I)$
C

$μ_{0} (N I)$
D

Zero

Solution

Here $\vec{B}$ (due to I) and $\vec{d l}$ will be in same direction.
So, $\oint \vec{B} . \vec{d l} = \oint B d l \cos 0^{0}$
$\oint B dl = \oint B dl = B N 2 πr = N (B 2 πr) = {Nμ}_{0} I$

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