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

The magnetic field will be
`\large{(1.4 * 10^(-4))/(d)} T`, '2d' being the distance the wires.

Step-by-step 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} = (\mu_(0)I)/(4 \pi)} \int \frac{\overrightarrow{dl} * \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}} = (\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) = (\mu_(0)I_(t))/(2 \pi d)`

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

`B_(b) = (\mu_(0)I_(b))/(2 \pi d)`

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

`\large{B_(M) = (4 \pi * 10^(-7))/(2 \pi d) (I_(t) - I_(b)) = (2 * 10^(-7))/(d) = (41.4 * 10 ^(-4))/(d)} T`