Question

A planar coil of area 7 $m^2$ carrying an anti-clockwise current 2 A is placed in an extemal magnetic field $\overrightarrow{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})$, such that the normal to the plane is along the line$(3\hat{i}-5\hat{j}+4\hat{k})$. Select correct statements from the following . ( Consider Normal of the coil and Magnetic moment vectors to be in the same direction )

• A. The potential energy of the coil in the given orientation is $6.4J$
• B. The angle between the normal (positive) to the coil and the external magnetic field is ${\cos }^{-1}\left(0.57\right)$
• C. The potential energy of the coil in the given orientatlon is $3.2J$
• D. The magnitude of magnetic moment of the coil is about 14 $A{m}^2$

Answer 1

Answer: (C), (D)

The correct options are
C The potential energy of the coil in the given orientatlon is $3.2J$
D The magnitude of magnetic moment of the coil is about 14 $A{m}^2$

Area $A=7m^2,$ Normal $\hat{n}=\frac{1}{\sqrt{50}}(3\hat{i}-5\hat{j}+4\hat{k})$
$\simeq \frac{1}{7}(3\hat{i}-5\hat{j}+4\hat{k})$
$\therefore$ area $\overrightarrow{A}=(3\hat{i}-5\hat{j}+4\hat{k})m^2$
Magnetic field $\overrightarrow{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})T$
Current $I=2A$
$\therefore$ magnetic moment $\overrightarrow{M}=I\overrightarrow{A}=(6\hat{i}-10\hat{j}+8\hat{k})A{m}^2$
$|\overrightarrow{M}|\simeq 14A{m}^2(2\times 7=14)$
Potential energy U$=-\overrightarrow{M}.\overrightarrow{B}$
$=-[1.2-2-2.4]=3.2J$
Let $\theta$ be the angle between $\hat{n}$ and $\overrightarrow{B}$(i.e.,)between $\overrightarrow{M}$ and $\overrightarrow{B}.$
Now,U$=-MBcos\theta .$
$\therefore cos\theta =-\frac{U}{MB}=-\frac{3.2}{14\times 0.4}(\because \sqrt{0.17}\simeq 0.4)$
$=-\frac{4}{7}=-0.57$
$\theta =\pi -co{s}^{-1}\left(0.57\right)$