Question 1: Magnetism and Matter

Question

A current carrying circular loop of radius 2 cm with unit normal $\hat{n}=\frac{\hat{k}+\hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B}=B_o(3 \hat{i}+2 \hat{k})$. If $B_o=4 	imes 10^{-3} \mathrm{~T}$ and current $I=100 \sqrt{2} \mathrm{~A}$, the torque experienced by the loop is $\_\_\_\_$ Wb.A. ( $\pi=3.14$ )

Accepted Answer

When a current-carrying loop is placed in an external magnetic field, it experiences a torque $(\vec{	au})$. The magnetic properties of the loop are represented by its magnetic dipole moment $(\overrightarrow{\mathrm{m}})$, defined as :
$$ \overrightarrow{\mathrm{m}}=\mathrm{I} \overrightarrow{\mathrm{~A}} $$
Where I is the current and $\vec{A}$ is the area vector of the loop. The direction of the area vector is perpendicular to the plane of the loop, given by the unit normal $\hat{\mathrm{n}}$.
$$ \overrightarrow{\mathrm{A}}=\mathrm{A} \hat{\mathrm{n}} $$
The torque is given by the cross product of the magnetic moment and the magnetic field :
$$ \vec{	au}=\overrightarrow{\mathrm{m}} 	imes \overrightarrow{\mathrm{B}} $$
$$\Rightarrow $$ $$ \vec{	au}=(\mathrm{IA} \hat{\mathrm{n}}) 	imes \overrightarrow{\mathrm{B}} $$
The radius of circular loop is $\mathrm{r}=2 \mathrm{~cm}=0.02 \mathrm{~m}$.
So, the area is
$$ \mathrm{A}=\pi \mathrm{r}^2=\pi(0.02)^2=4 \pi 	imes 10^{-4} \mathrm{~m}^2 $$
The unit normal is $\hat{\mathrm{n}}=\frac{\hat{\mathrm{k}}+\hat{\mathrm{t}}}{\sqrt{2}}$
And the current in loop is $\mathrm{I}=100 \sqrt{2} \mathrm{~A}$
So, the magnetic moment vector is;
$$ \overrightarrow{\mathrm{m}}=(100 \sqrt{2}) 	imes\left(4 \pi 	imes 10^{-4}ight) 	imes\left(\frac{\hat{\imath}+\hat{\mathrm{k}}}{\sqrt{2}}ight) $$
$$\Rightarrow $$ $$ \overrightarrow{\mathrm{m}}=400 \pi 	imes 10^{-4}(\hat{\imath}+\hat{\mathrm{k}})=0.04 \pi(\hat{\imath}+\hat{\mathrm{k}}) \mathrm{A} \cdot \mathrm{~m}^2 $$
The magnetic field in the region is $\vec{B}=B_o(3 \hat{\imath}+2 \hat{k})$ with $B_o=4 	imes 10^{-3} \mathrm{~T}$
$$ \overrightarrow{\mathrm{B}}=4 	imes 10^{-3}(3 \hat{\imath}+2 \hat{\mathrm{k}})=(12 \hat{\imath}+8 \hat{\mathrm{k}}) 	imes 10^{-3} \mathrm{~T} $$
Using the formula for magnetic torque,
$$ \vec{	au}=\overrightarrow{\mathrm{m}} 	imes \overrightarrow{\mathrm{B}} $$
$$\Rightarrow $$ $\vec{	au}=[0.04 \pi(\hat{\imath}+\hat{k})] 	imes\left[(12 \hat{\imath}+8 \hat{k}) 	imes 10^{-3}ight]$
$$\Rightarrow $$ $\vec{	au}=\left(0.04 \pi 	imes 10^{-3}ight)[(\hat{\imath}+\hat{k}) 	imes(12 \hat{\imath}+8 \hat{k})]$
$$\Rightarrow $$ $\vec{	au}=\left(0.04 \pi 	imes 10^{-3}ight)(4 \hat{\jmath})$
$$\Rightarrow $$ $\vec{	au}=0.16 \pi 	imes 10^{-3} \hat{\jmath}$
$$\Rightarrow $$ $\vec{	au}=0.16 	imes 3.14 	imes 10^{-3} \hat{\jmath}=0.5024 	imes 10^{-3} \hat{\jmath}$
$$\Rightarrow $$ $$ \vec{	au}=5024 	imes 10^{-7} \hat{\jmath} $$
Therefore, the torque experienced by the loop is $5024 	imes 10^{-7} \hat{\mathrm{j}}$.
Hence, the correct option is (D).

Answer

$$ 16 	imes 10^{-5} \hat{k} $$

Answer

$$ 5024 	imes 10^{-7} \hat{k} $$

Answer

$$ 5024 	imes 10^{-7} \hat{i} $$

Answer

$$ 5024 	imes 10^{-7} \hat{j} $$

Magnetism and Matter

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