Question: Q. 3. A current carrying circular loop of radius $R$ is placed in the $x-y$ plane with centre at the origin. Half of the loop with $x>0$ is now bent so that it now lies in the $y-z$ plane. (a) The magnitude of magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of $B$ at $(0,0, z), z»R$ increases.
(d) The magnitude of $B$ at $(0,0, z), z»R$ is unchanged.
[NCERT Exemplar]
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Solution:
Ans. Correct option : (a)
Explanation: For a circular loop of radius $R$, carrying current $I$ in $x-y$ plane, the magnetic moment
$$ M=I \times \pi R_{2} $$
It acts perpendicular to the loop along $z$-direction.
When half of the eurrent loop is bent in $y-z$ plane, then magnetic moment due to half current loop is $x-y$ plane, $M_{1} \in I\left(x R_{2} / 2\right)$ acting along $z$-direction.
Magnetic moment due to half current loop in $y-z$ plane, $\mathrm{M}{2} \Rightarrow I\left(\pi R{2} / 2\right)$ along $x$-direction.
Net-magnetic moment due to entire bent current loop
$$ \begin{aligned} M_{n e t} & =\sqrt{M_{1}^{2}+M_{2}^{2}} \ & =\sqrt{2} \frac{I \pi R^{2}}{2} \ & =\frac{M}{\sqrt{2}} \end{aligned} $$
Therefore, $M_{\text {net }}<M$ or $M$ diminishes.
