The coil in Figure carries current i=2.00Ain the direction indicated, is parallel to an xz plane, has 3.00turns and an area of $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{m}}^{\mathbf{2}}$, and lies in a uniform magnetic field $\mathbf{B}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}\hat{\mathbf{k}}\mathbf{)}\mathbf{mT}$(a) What are the orientation energy of the coil in the magnetic field (b)What are the torque (in unit-vector notation) on the coil due to the magnetic field?

1. The current through the coil, i= 2.00A
2. The number of turns, N=3.00
3. The area of coil,$A=4.00\times {10}^{-3}{m}^2$
4. The uniform magnetic field,$\begin{array}{l}{}^B=(2.00\hat{i}-3.00\stackrel{⏜}{]}-4.00\hat{k}))\mathrm{mT}\\ =(2.00\mathrm{i}-3.00\hat{j}-4.00\hat{k})\times {10}^{-3}T\end{array}$

By takingthe scalar product of the magnetic dipole moment $\overrightarrow{\mu }$of the coil with the magnetic field $\overrightarrow{B}$, we can find the orientation energy of the coil. By taking vector product ofthe magnetic dipole moment $\overrightarrow{\mu }$and magnetic field$\overrightarrow{B}$, we can find thetorque on the coil due to the magnetic field.

Formula:

1. The orientation energy of the magnetic dipole is$U=-\mu ⃗.B⃗$
2. The magnitude of$\overrightarrow{\mu }$is$\mu =NiA$
3. Torque on the coil is$\tau ⃗=\mu ⃗\times B⃗$

The orientation energy of the magnetic dipole is given by

$U=-\mu ⃗.B⃗$

Where $\overrightarrow{\mu }$is the magnetic dipole moment of the coil and $\overrightarrow{B}$is the magnetic field

We know that the magnitude of $\overrightarrow{\mu }$is

$\mu =NiA$

Where i is the current in the coil, N is the number of turns, A is the area of coil.

By using the right-hand rule, we see that $\overrightarrow{\mu }$is in -y direction.

Thus, we have,

$\mu =NiA(-\hat{j})$

$\begin{array}{l}\mu =(-3\times 2.00A\times 4.00\times {10}^{-3}{m}^2)\hat{j}\\ =(-0.0240A{m}^2)\hat{j}\end{array}$

The corresponding orientation energy is given by

role="math" localid="1662967936394" $$\begin{gathered} U=-\mu ⃗.B⃗ \\ U=-{\mu }_y{B}_y \end{gathered}$$

U= (-0.0240A.m²)$\times (-3.00\times {10}^{-3}T)$

$=-7.20\times {10}^{-5}J$

The torque on the coil is given by

$\tau ⃗=\mu ⃗\times B⃗$

Since , $\hat{j}.\hat{i}=0,\hat{j}\times \hat{j}=0and\hat{j}\times \hat{k}=\hat{i}$

Therefore, the torque on the coil is

$\begin{array}{l}\tau ={\mu }_y{B}_z\hat{i}-{\mu }_y{B}_x\hat{k}\\ =(-0.024\mathrm{A}.{\mathrm{m}}^2\left)\right((4.00\times {10}^{-3}T)\hat{i}-(2.00x{10}^{-3}T)\hat{k}\\ =(9.60\times {10}^{-5}Nm)\hat{i}=+(4.80\times {10}^{-5}Nm)\hat{k}\end{array}$