Question

A wire carrying current $I$ is bent in the shape$ABCDEFA$ as shown, where rectangle $ABCDA$ and $ADEFA$ are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$, then the magnitude and direction of magnetic moment of the loop $ABCDEFA$ is:

• A. $\sqrt{2}abI,$along $\left[\frac{{\displaystyle \stackrel{\wedge }{j}}}{\sqrt{5}}+\frac{{\displaystyle 2\stackrel{\wedge }{k}}}{\sqrt{5}}\right]$
• B. $abI,$along $\left[\frac{{\displaystyle \stackrel{\wedge }{j}}}{\sqrt{5}}+\frac{{\displaystyle 2\stackrel{\wedge }{k}}}{\sqrt{5}}\right]$
• C. $\sqrt{2}abI,$along $\left[\frac{{\displaystyle \stackrel{\wedge }{j}}}{\sqrt{2}}+\frac{{\displaystyle \stackrel{\wedge }{k}}}{\sqrt{2}}\right]$
• D. $abI,$along $\left[\frac{{\displaystyle \stackrel{\wedge }{j}}}{\sqrt{2}}+\frac{{\displaystyle \stackrel{\wedge }{k}}}{\sqrt{2}}\right]$

Answer 1

The correct option is C

$\sqrt{2}abI,$along $\left[\frac{{\displaystyle \stackrel{\wedge }{j}}}{\sqrt{2}}+\frac{{\displaystyle \stackrel{\wedge }{k}}}{\sqrt{2}}\right]$

Step 1. Finding a magnetic moment of loop $ABCD$.

From, Loop ABCD in the diagram, the magnetic moment ${\overrightarrow{M}}_1$ is passing in the $k-$direction is given below,

${\overrightarrow{M}}_1=\left(abI\right)\stackrel{\wedge }{k}$

Where ${\overrightarrow{M}}_1$ is the magnetic moment

$a$ is the length between $A$ and $B$

$b$is the length between $B$ and $C$

$I$ is the current passing in the loop

Step 2. Finding a magnetic moment of loop $DEFA$.

Similarly, in Loop $DEFA$, the magnetic moment ${\overrightarrow{M}}_2$ is passing in the $j-$direction is given below,

${\overrightarrow{M}}_2=\left(abI\right)\stackrel{\wedge }{j}$

Where ${\overrightarrow{M}}_2$ is the magnetic moment

$a$ is the length between $A$ and $B$

$b$is the length between $B$ and $C$

$I$ is the current passing in the loop

Step 3. Calculate the magnetic moment of loop $ABCDEFA$.

Now, $$\begin{gathered} \overrightarrow{M}={\overrightarrow{M}}_1+{\overrightarrow{M}}_2=abI\left(\stackrel{\wedge }{j}+\stackrel{\wedge }{k}\right) \\ \left|\overrightarrow{M}\right|=\sqrt{2}abI \end{gathered}$$

Direction, along $\left[\frac{{\displaystyle \stackrel{\wedge }{j}}}{\sqrt{2}}+\frac{{\displaystyle \stackrel{\wedge }{k}}}{\sqrt{2}}\right]$

Therefore, the correct option is (C).