Ncert Exercises & Solutions

Two identical current-carrying coaxial loops carry current

I

$I$ in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as

C,

$C,$

(a)	$\oint B\cdot dl= \mp 2\mu_0 I$
(b)	the value of $\oint B\cdot dl$ is independent of the sense of $C$ .
(c)	there may be a point on $C$ where $B$ and $dl$ are perpendicular.
(d)	$B$ vanishes everywhere on $C$ .

Which of the above statements is correct?

1.	(a) and (b)	2.	(a) and (c)
3.	(b) and (c)	4.	(c) and (d)

Hint: The magnetic field depends on the current.

Explanation: Consider a simple amperian loop passing once through both the identical current-carrying coaxial loops.
(i) According to Ampere circuital law,

\oint_{C} \to B \cdot d l = μ_{0} (I - I) = 0.

$\oint_C \vec{B} \cdot d l=\mu_0(I-I)=0.$ Hence, option (a) is wrong.
(ii) As

\oint_{C} \to B \cdot d l = 0,

$\oint_C \vec{B} \cdot d l=0,$ therefore

\oint_{C} \to B \cdot d l

$\oint_C \vec{B} \cdot d l$ is independent of the sense of

C .

$C.$ Thus option (b) is correct.
(iii) There will be a point on loop

C,

$C,$ lying at the axis of two loops

A

$A$ and

B,

$B,$ where

\to B

$\vec{B}$ and

\to d

$\vec{d}$ are perpendicular to each other. Thus option (c) is correct.
(iv) The value of

\to B

$\vec{B}$ does not vanish on various points of

C .

$C.$

Hence, option (3) is the correct answer.

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