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NCERT Exemplar · Q8

Q.Two identical current carrying coaxial loops, carry current II in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as CC,

(a) ∮cB⃗⋅dl⃗=∓2μ0I\oint_c \vec{B}\cdot d\vec{l} = \mp 2\mu_0 I.
(b) the value of ∮cB⃗⋅dl⃗\oint_c \vec{B}\cdot d\vec{l} is independent of sense of C.
(c) there may be a point on C where B and dl are perpendicular.
(d) B vanishes everywhere on C.
Yanam BieapMCQ· 1mImportance★★★★★est
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The net current enclosed by CC is zero for either sense of traversal, so ∮CB⃗⋅dl⃗=0\oint_C \vec{B}\cdot d\vec{l} = 0 regardless of the sense of CC - matching options (b) and (c).

Setting up Ampere's law

∮CB⃗⋅dl⃗=μ0Ienc\oint_C \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}

The two coaxial loops carry equal current II but in opposite senses. The amperian loop CC threads through both of them exactly once, so

Ienc=I−I=0⇒∮CB⃗⋅dl⃗=0.I_{\text{enc}} = I - I = 0 \qquad\Rightarrow\qquad \oint_C \vec{B}\cdot d\vec{l} = 0.

Checking (a)

(a) claims ∮CB⃗⋅dl⃗=∓2μ0I\oint_C \vec{B}\cdot d\vec{l} = \mp 2\mu_0 I. This double-counts the current as if both loops contributed with the same sign; since they are opposite, the terms cancel to zero, not 2μ0I2\mu_0 I. (a) is false.

Checking (b)

Reversing the sense in which CC is traversed flips the sign of IencI_{\text{enc}} (from I−II-I to −(I−I)-(I-I)), but since Ienc=0I_{\text{enc}}=0 either way, the value of the line integral is 00 regardless of which sense CC is traversed. (b) is true.

Checking (c) …

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