Question

A circular loop of wire having a radius of 8.0cm carries a current of 0.20A. A vector of unit length and parallel to the dipole moment$\overrightarrow{\mathbf{\mu }}$of the loop is given by $\mathbf{0}\mathbf{.}\mathbf{60}\hat{\mathbf{i}}\mathbf{}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{80}\hat{\mathbf{j}}$. (This unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in a uniform magnetic field given by$\mathbf{B}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{T}\mathbf{)}\mathbf{}\hat{\mathbf{i}}\mathbf{}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{T}\mathbf{)}\mathbf{}\hat{\mathbf{k}}$(a)Find the torque on the loop (in unit vector notation)(b)Find the orientation energy of the loop.

Short answer

1. The torque acting on the loop is $(-9.7\times {10}^{-4})\hat{i}-(7.2\times {10}^{-4})\hat{j}+(8.0\times {10}^{-4})\hat{k}N\cdot m$
2. The orientation energy of the loop is -6.0*10-4J.

Step-by-step solution

Identification of given data

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

Significance of torque and orientation energy of a magnetic field

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

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

Formula:

role="math" localid="1662971704358" $\tau ⃗=\mu ⃗\times B⃗$ ...(i)

Here, $\overrightarrow{\tau }$is the torque experienced by the current carrying coil, role="math" localid="1662970793088" $\overrightarrow{\mu }$is the magnetic dipole moment, and $\overrightarrow{B}$is the magnetic field.

$U\left(\theta \right)=-\mu ⃗\cdot B⃗$ ...(ii)

Here, U$\left(\theta \right)$is the orientation energy, $\overrightarrow{\mu }$is the magnetic dipole moment, and $\overrightarrow{B}$is the magnetic field. $\mu =NiA$ ...(iii)

Here, $\mu$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.

Determining the value of μ

The magnetic dipole moment is calculated as,

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

The value of $\mu$is calculated using equation (i) as follows:

role="math" localid="1662971327706" $$\begin{gathered} \mu =NiA \\ =Ni{\mathrm{\pi r}}^2 \end{gathered}$$

Substitute the values of current, radius and number of turns to calculate $\mu$.

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

Therefore, the value of $\mu$is 4.02$\times {10}^{-3}A.m^2$

(a) Determining the torque

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

$$\begin{gathered} \tau ⃗=\mu ⃗\times B⃗ \\ =\mu \left(0.60\hat{i}-0.80\hat{j}\right)\times \left(\left(0.25T\right)\hat{i}+\left(0.30T\right)\hat{k}\right) \\ =\mu (\left(\left(0.60\right)\left(0.30\right)\left(\hat{i}\times \hat{j}\right)\right)-(\left(0.80\right)\left(0.25\right)(\hat{j}\times \hat{i}))-((0.80)(0.30)(\hat{j}\times \hat{k})\left)\right)T \\ =\mu (-0.24\hat{i}-0.18\hat{j}+0.20\hat{k})T \end{gathered}$$

Now, substitute the value of $\mu$calculated in step 1.

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

Therefore, the torque acting on the loop is,

$(-9.7\times {10}^{-4})\hat{i}-(7.2\times {10}^{-4})\hat{j}+(8.0\times {10}^{-4})\hat{k}N\cdot m$

(b) Determining the orientation energy

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

$$\begin{gathered} U\left(\theta \right)=-\mu (0.60\hat{i}-0.80\overbrace{j}).\left(\left(0.25T\right)\hat{i}+(0.30T)\hat{k}\right) \\ =-\mu (0.60)(0.25)T \\ =-0.15\mu \\ =-6.0\times {10}^{-4}J \end{gathered}$$

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