91Ó°ÊÓ

The magnetic field created by the electric field equals the size of that electric current with the same proportions as the free access space.

The magnitude of the magnetic field due to current-carrying wire is,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{Ï€°ù}}$

Here, B is the magnetic field, ${\mathbf{μ}}_{\mathbf{0}}$is the permeability of free space, I is the current, and r is the radius.

(a) Determination of the magnitude and direction of the current ${\mathbf{I}}_{\mathbf{2}}$to get zero net field at P.

Determine the directions of the fields at point P due to the wires from the image below.

The resultant field at point P is zero and this is possible if the magnetic fields due to the current carrying wires, i.e. ${\stackrel{→}{\mathrm{B}}}_1$and ${\stackrel{→}{\mathrm{B}}}_2$are equal and opposite in direction. From the figure, as is pointing upwards so the current is pointing out of the page

$\begin{array}{r}{\mathrm{B}}_1=\frac{{\mathrm{μ}}_0{\mathrm{l}}_1}{2{\mathrm{Ï€°ù}}_1}\\ =\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\left(\frac{6.00\mathrm{A}}{1.50\mathrm{m}}\right)\\ {\mathrm{B}}_2=\frac{{\mathrm{μ}}_0{\mathrm{I}}_2}{2{\mathrm{Ï€°ù}}_2}\\ =\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\left(\frac{{\mathrm{I}}_2}{0.50\mathrm{m}}\right)\\ {\mathrm{B}}_1={\mathrm{B}}_2\\ \frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\left(\frac{6.00\mathrm{A}}{1.50\mathrm{m}}\right)=\frac{{\mathrm{μ}}_0}{2\mathrm{Ï€}}\left(\frac{{\mathrm{I}}_2}{0.50\mathrm{m}}\right)\\ {\mathrm{I}}_2=\left(\frac{0.50\mathrm{m}}{1.50\mathrm{m}}\right)(6.00\mathrm{A})\\ =2.00\mathrm{A}\end{array}$

So, the magnitude and direction of the current ${\mathrm{I}}_2$is 2.00 A out of the page.

For point Q, the directions of the fields are depicted in the figure below,

Calculate the fields by substituting all the values,

$\begin{array}{r}{\mathrm{B}}_1=\frac{{\mathrm{μ}}_0{\mathrm{I}}_1}{2{\mathrm{Ï€°ù}}_1}\\ =\left(2×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(\frac{6.00\mathrm{A}}{0.50\mathrm{m}}\right)\\ =2.40×{10}^{−6}\mathrm{T}\\ {\mathrm{B}}_2=\frac{{\mathrm{μ}}_0{\mathrm{I}}_2}{2{\mathrm{Ï€°ù}}_2}\\ =\left(2×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(\frac{2.00\mathrm{A}}{1.50\mathrm{m}}\right)\\ =2.67×{10}^{−7}\mathrm{T}\end{array}$

So, ${\mathrm{B}}_{1-}{\mathrm{B}}_2$ are opposite to each other and from the calculated values,localid="1668339938597" ${\mathrm{B}}_1>{\mathrm{B}}_2$.

Therefore, the resultant magnetic field is,

$\begin{array}{r}\mathrm{B}={\mathrm{B}}_1−{\mathrm{B}}_2\\ =\left(2.40×{10}^{−6}\mathrm{T}\right)−\left(2.67×{10}^{−7}\mathrm{T}\right)\end{array}$

Hence, the magnitude of the net field at Q is $2.13×{10}^{−6}\mathrm{T}$in the upward direction.

The image below depicts the vectors that describe the dynamics at point S,

Calculate the fields by substituting all the values,

$$\begin{gathered} {\mathrm{B}}_1=\frac{{\mathrm{μ}}_0{\mathrm{l}}_1}{2\mathrm{Ï€}{\mathrm{r}}_1} \\ \left.=\left(2×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\right)\left(\frac{6.00\mathrm{A}}{0.60\mathrm{m}}\right) \\ =2.00×{10}^{−6}\mathrm{T} \end{gathered}$$

$\begin{array}{r}{\mathrm{B}}_2=\frac{{\mathrm{μ}}_0{\mathrm{I}}_2}{2\mathrm{Ï€}{\mathrm{r}}_2}\\ \left.=\left(2×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\right)\left(\frac{2.00\mathrm{A}}{0.80\mathrm{m}}\right)\\ =5.00×{10}^{−7}\mathrm{T}\end{array}$

At S, the fields are at right angle to each other. So, their resultant can be determined by the vector formula of magnitudes.

localid="1668339947842" $$\begin{gathered} B=\sqrt{B_1^2+B_2^2} \\ =\sqrt{{\left(2.00×{10}^{−6}\mathrm{T}\right)}^2+{\left(5.00×{10}^{−7}\mathrm{T}\right)}^2} \\ =2.06×{10}^{−6}\mathrm{T} \end{gathered}$$

Thus, the resultant magnetic field at point S islocalid="1668339952305" $2.06×{10}^{−6}\mathrm{T}$.