91Ó°ÊÓ

1. Radius of the wire, r = 8.00cm
2. Current in the wire, i = 0.20A
3. The dipole moment is,$\mathrm{μ}⃗=0.60\hat{i}-0.80\hat{j}$
4. The magnetic field is, localid="1663952855032" $B⃗=(0.25T)\hat{i}+(0.30T)\hat{k}$

The torque, $\stackrel{→}{\mathrm{τ}}$experienced by the current carrying coil of area A and turns N, and carrying a current i in the uniform magnetic field $\stackrel{→}{B}$is equal to the cross product of magnetic dipole moment$\stackrel{→}{\mathrm{μ}}$ of the coil and magnetic field $\stackrel{→}{B}$.

The orientation energy of a magnetic dipole in a magnetic field is equal to the negative dot product of the magnetic dipole moment $\stackrel{→}{\mathrm{μ}}$of the coil and magnetic field $\stackrel{→}{B}$.

Formula:

role="math" localid="1662971704358" $\mathrm{τ}⃗=\mathrm{μ}⃗×B⃗$ ...(i)

Here, $\stackrel{→}{\mathrm{τ}}$is the torque experienced by the current carrying coil, role="math" localid="1662970793088" $\stackrel{→}{\mathrm{μ}}$is the magnetic dipole moment, and $\stackrel{→}{B}$is the magnetic field.

$U\left(\mathrm{θ}\right)=-\mathrm{μ}⃗â‹\dots B⃗$ ...(ii)

Here, U$\left(\mathrm{θ}\right)$is the orientation energy, $\stackrel{→}{\mathrm{μ}}$is the magnetic dipole moment, and $\stackrel{→}{B}$is the magnetic field. $\mathrm{μ}=NiA$ ...(iii)

Here, $\mathrm{μ}$is the magnetic dipole moment, N is the number of turns of the coil, i is current in the coil, and A is the area of the cross-section of the coil.

The magnetic dipole moment is calculated as,

$\mathrm{μ}⃗=\mathrm{μ}(0.60\hat{i}-0.80\hat{j}).$

The value of $\mathrm{μ}$is calculated using equation (i) as follows:

role="math" localid="1662971327706" $$\begin{gathered} \mathrm{μ}=NiA \\ =Ni{\mathrm{Ï€°ù}}^2 \end{gathered}$$

Substitute the values of current, radius and number of turns to calculate $\mathrm{μ}$.

$$\begin{gathered} \mathrm{μ}=1(0.20A)\mathrm{π}{\left(0.080\mathrm{m}\right)}^2 \\ =4.02×{10}^{-3}\mathrm{A}.{\mathrm{m}}^2 \end{gathered}$$

Therefore, the value of $\mathrm{μ}$is 4.02$×{10}^{-3}A.m^2$

Calculate the value of torque using the equation (i) as below:

$$\begin{gathered} \mathrm{τ}⃗=\mathrm{μ}⃗×B⃗ \\ =\mathrm{μ}\left(0.60\hat{i}-0.80\hat{j}\right)×\left(\left(0.25T\right)\hat{i}+\left(0.30T\right)\hat{k}\right) \\ =\mathrm{μ}(\left(\left(0.60\right)\left(0.30\right)\left(\hat{i}×\hat{j}\right)\right)-(\left(0.80\right)\left(0.25\right)(\hat{j}×\hat{i}))-((0.80)(0.30)(\hat{j}×\hat{k})\left)\right)T \\ =\mathrm{μ}(-0.24\hat{i}-0.18\hat{j}+0.20\hat{k})T \end{gathered}$$

Now, substitute the value of $\mathrm{μ}$calculated in step 1.

$$\begin{gathered} \stackrel{→}{\mathrm{τ}}=4.02×{10}^{-3}A.m^2(-0.24\hat{i}-0.18\hat{j}+0.20\hat{k}) \\ =(-9.7×{10}^{-4})\hat{i}-(7.2×{10}^{-4})\hat{j}+(8.0×{10}^{-4})\hat{k}N.m \end{gathered}$$

Therefore, the torque acting on the loop is,

$(-9.7×{10}^{-4})\hat{i}-(7.2×{10}^{-4})\hat{j}+(8.0×{10}^{-4})\hat{k}Nâ‹\dots m$

Use the equation (iii) to calculate the orientation energy.

$$\begin{gathered} U\left(\mathrm{θ}\right)=-\mathrm{μ}(0.60\hat{i}-0.80\stackrel{Áåž}{j}).\left(\left(0.25T\right)\hat{i}+(0.30T)\hat{k}\right) \\ =-\mathrm{μ}(0.60)(0.25)T \\ =-0.15\mathrm{μ} \\ =-6.0×{10}^{-4}J \end{gathered}$$

JTherefore, the orientation energy of the loop is $-6.0×{10}^{-4}$.