91Ó°ÊÓ

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

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

Formula:

1. The orientation energy of the magnetic dipole is$U=-\mathrm{μ}⃗.B⃗$
2. The magnitude of$\stackrel{→}{\mathrm{μ}}$is$\mathrm{μ}=NiA$
3. Torque on the coil is$\mathrm{τ}⃗=\mathrm{μ}⃗×B⃗$

The orientation energy of the magnetic dipole is given by

$U=-\mathrm{μ}⃗.B⃗$

Where $\stackrel{→}{\mathrm{μ}}$is the magnetic dipole moment of the coil and $\stackrel{→}{B}$is the magnetic field

We know that the magnitude of $\stackrel{→}{\mathrm{μ}}$is

$\mathrm{μ}=NiA$

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 $\stackrel{→}{\mathrm{μ}}$is in -y direction.

Thus, we have,

$\mathrm{μ}=NiA(-\hat{j})$

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

The corresponding orientation energy is given by

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

U= (-0.0240A.m²)$×(-3.00×{10}^{-3}T)$

$=-7.20×{10}^{-5}J$

The torque on the coil is given by

$\mathrm{τ}⃗=\mathrm{μ}⃗×B⃗$

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

Therefore, the torque on the coil is

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