Question

Two long, straight, parallel wires, 10.0 cm apart, carry equal 4.00 A currents in the same direction, as shown in Fig. Find the magnitude and direction of the magnetic field at point${\mathit{P}}_{\mathbf{3}}$, 20.0 cm directly above${\mathit{P}}_{\mathbf{1}}$.

Short answer

The magnetic field at point is $7.54\mu T$in left direction.

Step-by-step solution

Step 1:  The magnetic field due to wire at a point

The magnetic field due to wire at point is given by

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\pi r}}$

Where, B is the magnetic field due to wire,${\mu }_0$is the permeability of vaccum, l is the current through the wire and r is the distance from wire.

The total magnetic field at point due two wires

The magnetic field at point due two wires is given by vector summation of the magnetic field of two wires at that point.

${\overrightarrow{\mathrm{B}}}_{\mathrm{T}}={\overrightarrow{\mathrm{B}}}_{1\mathrm{x}}+{\overrightarrow{\mathrm{B}}}_{2x}$

Where, ${\overrightarrow{B}}_T$is the total magnetic field, ${\overrightarrow{\mathrm{B}}}_{1x}\text{and}{\overrightarrow{B}}_{2x}$are magnetic fields due to first wire and second wire.

Calculation of distance of point P3 from wire 1  and wire 2 .

Given : The current carried by both wire is l = 4.00A

The distance between two wires is 20.0cm

Using Pythagoras theorem

_{$\begin{array}{r}\left({\text{Hypotenuse}}^2\right.=(\text{perpendicular}{)}^2+(\text{Base}{)}^2\\ r^2=(20.0\mathrm{cm}{)}^2+(5.00\mathrm{cm}{)}^2\\ r=\sqrt{400+25}\\ r=20.6\mathrm{cm}\end{array}$}

Where, r is the distance of point $P_3$from wire 1 and wire 2

Calculation of the total magnetic field at point due two wires 

Using

$\begin{array}{r}{\overrightarrow{B}}_T={\overrightarrow{B}}_{1x}+{\overrightarrow{B}}_{2x}\\ {\overrightarrow{B}}_T=-2\frac{{\mu }_{0}I}{2\pi r}\cos \theta \end{array}$

_{Where , $\theta$ is the angle between of magnetic field vector and horizontal.}

Now, putting the values of constants in above equation

_{$\begin{array}{r}{\overrightarrow{B}}_T=-2\frac{{\mu }_{0}l}{2\pi r}\cos \theta \widehat{\mathrm{i}}\\ {\overrightarrow{B}}_T=-\left(\frac{4\times {10}^{-7}\times 4}{20.6\times {10}^{-2}}\times \frac{0.20}{20.6\times {10}^{-2}}\right)\widehat{\mathrm{i}}\\ {\overrightarrow{B}}_T=-\left(7.54\mu \mathrm{T}\right)\widehat{\mathrm{i}}\end{array}$}

Thus, the magnetic field at point is $7.45\mu T$in left direction .