Question

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bottom one carries 12.5 A ? Express your answer using three significant figures.

Answer 1

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ($\large{\overrightarrow{B}}$) at a distance '$r$' due to a current carrying conductor carrying current '$I$' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where '$\overrightarrow{dl}$' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current '$I$', at a distance '$d$' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if '$I_{t}$' be the current carried by the top wire, '$I_{b}$' be the current carried by the bottom wire and '$2d$' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ($\large{B_{t}}$) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ($\large{B_{b}}$) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ($\large{B_{M}}$) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$