Question

The resultant magnetic moment for the following arrangement is?

• A. 4M
• B. 3M
• C. 2M
• D. M

Answer 1

The correct option is C

2M

Step 1: Given that:

The vectors in the given arrangement can be resolved as -

$-M\overrightarrow{i}$

$M\cos \left(-60°\right)\overrightarrow{i}+M\sin \left(-60°\right)\overrightarrow{j}$

here,

$\overrightarrow{i}$ is the unit vector in + x-direction.

$\overrightarrow{j}$is the unit vector in + y-direction

$-M\overrightarrow{i}$

$M\cos \left(-60°\right)\overrightarrow{i}+M\sin \left(-60°\right)\overrightarrow{j}$

The 4 vectors can be given by arrangement

Step 2: Formula used

$$\begin{gathered} \overrightarrow{r}=a\overrightarrow{i}+b\overrightarrow{j} \\ \left|\overrightarrow{r}\right|=\sqrt{a^2+b^2} \end{gathered}$$

Step 3: Calculating The magnitude of the net magnetic moment

Since the magnetic moment is a vector quantity it follows the vector law of addition

adding 1+2+3+4

$$\begin{gathered} M_{net}=\left(-2M+2M\cos (-60°)\right)\overrightarrow{i}+\left(2M\sin (-60°)\right)\overrightarrow{j} \\ {M}_{net}=\left(-2M+2M\times \frac{1}{2}\right)\overrightarrow{i}+(2M\times \frac{-\sqrt{3}}{2})\overrightarrow{j} \\ {M}_{net}=-M\overrightarrow{i}-\sqrt{3}M\overrightarrow{j} \\ \left|M_{net}\right|=\sqrt{{(-M)}^2+{(-\sqrt{3}M)}^2} \\ \left|M_{net}\right|=\sqrt{M^2+{3M}^2} \\ \left|M_{net}\right|=\sqrt{4M^2} \\ \left|M_{net}\right|=2M \end{gathered}$$

Hence, option C is correct