Question

A circular loop of wire with area A lies in the xy-plane. As viewed along the z-axis looking in the -z direction toward the origin, a current I is circulating clockwise around the loop. The torque produced by an external magnetic field $\overrightarrow{\mathbf{B}}$is given by $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{\tau }}\mathbf{=}\mathbf{D}\left(4\hat{i}-3\hat{j}\right)$, where is a positive constant, and for this orientation of the loop the magnetic potential energy $\mathbf{U}\mathbf{=}\mathbf{-}\mathbf{\mu }\mathbf{.}\overrightarrow{\mathbf{B}}$Isnegative. The magnitude of the magnetic field is${\mathbf{B}}_{\mathbf{\circ }}\mathbf{=}\mathbf{13}\mathbf{D}\mathbf{/}\mathbf{IA}$

(a) Determine the vector magnetic moment of the current loop.

(b) Determine the components${\mathbf{B}}_{\mathbf{x}}\mathbf{,}{\mathbf{B}}_{\mathbf{y}}\mathbf{\text{and}}{\mathbf{B}}_{\mathbf{z}}\mathbf{\text{of}}\mathbf{B}$

Short answer

The vector magnetic moment of the current loop is and the components

${\mathrm{B}}_{\mathrm{x}}=\frac{3\mathrm{D}}{\mathrm{IA}},{\mathrm{B}}_{\mathrm{y}}=\frac{4\mathrm{D}}{\mathrm{IA}}\text{and}{\mathrm{B}}_{\mathrm{z}}=\frac{-12\mathrm{D}}{\mathrm{IA}}\text{of}\mathrm{B}$.

Step-by-step solution

Definition of Magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the vector magnetic moment of the current loop and components Bx,By and Bz of B

Using right hand rule the vector magnetic moment of the current loop is calculated as

$\begin{array}{r}\overrightarrow{\mathrm{\mu }}=-\mathrm{\mu }\widehat{\mathrm{k}}\\ \overrightarrow{\mathrm{\mu }}=-\mathrm{IA}\widehat{\mathrm{k}}\end{array}$

Hence, the vector magnetic moment of the current loop is

Now the components can be calculated as

$\overrightarrow{\mathrm{B}}={\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{k}}$

And the torque$\overrightarrow{\mathrm{T}}=\overrightarrow{\mathrm{\mu }}\times \overrightarrow{\mathrm{B}}$

Than

$\begin{array}{r}\overrightarrow{\mathrm{T}}=\overrightarrow{\mathrm{\mu }}\times \overrightarrow{\mathrm{B}}\\ \overrightarrow{\mathrm{T}}=(-\mathrm{IA})\left({\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{k}}\times \widehat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{k}}\times \widehat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{k}}\times \widehat{\mathrm{k}}\right)\\ \overrightarrow{\mathrm{T}}={\mathrm{IAB}}_{\mathrm{y}}\widehat{\mathrm{i}}-{\mathrm{IAB}}_{\mathrm{x}}\widehat{\mathrm{j}}\end{array}$

And the torque obtain by external magnetic field$\overrightarrow{\mathrm{T}}=\mathrm{D}(4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}})$

Compare both of them get but not

${\mathrm{B}}_{\mathrm{y}}=\frac{4\mathrm{D}}{1\mathrm{A}},{\mathrm{B}}_{\mathrm{x}}=\frac{3\mathrm{D}}{1\mathrm{A}}$

Now

$\begin{array}{r}{\mathrm{B}}_0^2={\mathrm{B}}_{\mathrm{x}}^2+{\mathrm{B}}_{\mathrm{y}}^2+{\mathrm{B}}_{\mathrm{z}}^2\\ {\mathrm{B}}_{\mathrm{z}}=\pm \sqrt{{\mathrm{B}}_0^2-{\mathrm{B}}_{\mathrm{x}}^2-{\mathrm{B}}_{\mathrm{y}}^2}\end{array}$

Put all values

$\begin{array}{r}{\mathrm{B}}_{\mathrm{z}}=\pm \left(\frac{\mathrm{D}}{\mathrm{IA}}\right)\sqrt{169-9-16}\\ {\mathrm{B}}_{\mathrm{z}}=\pm 12\frac{\mathrm{D}}{\mathrm{IA}}\end{array}$

For sign of${\mathrm{B}}_{\mathrm{z}}$use U

$\begin{array}{r}\mathrm{U}=-\overrightarrow{\mathrm{\mu }}\cdot \overrightarrow{\mathrm{B}}\\ =-(-\mid \mathrm{A}\widehat{\mathrm{k}})\cdot \left({\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{k}}\right)\\ \mathrm{U}=+\mid {\mathrm{AB}}_{\mathrm{z}}\end{array}$

It means U is negative so${\mathrm{B}}_{\mathrm{z}}$is negative

Than ${\mathrm{B}}_{\mathrm{z}}=\frac{-12\mathrm{D}}{\mathrm{IA}}$

Hence, the components ${\mathrm{B}}_{\mathrm{x}}=\frac{3\mathrm{D}}{\mathrm{IA}},{\mathrm{B}}_{\mathrm{y}}=\frac{4\mathrm{D}}{\mathrm{IA}}\text{and}{\mathrm{B}}_{\mathrm{z}}=\frac{-12\mathrm{D}}{\mathrm{IA}}\text{of}\mathrm{B}$.