Question

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is the magnitude of the magnetic field at a point that is 5.0 cm from each of the wires

Answer 1

The magnitude of magnetic field at given point =
`5.33` ×
`10^(-5)` T

Step-by-step explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒
`B = (\mu_(0)i )/(2\pi R)`

Where
`B =` magnetic field due to long wires,
`\mu_(0) =`
`4\pi *10^(-7)`,
`R =` perpendicular distance from wire to given point

From any one wire
`R_(1) =` 5 cm,
`R_(2) =` 3 cm

so we write,

∴
`B = B_(1) + B_(2)`

`B = (\mu_(0) i)/(2\pi R_(1) ) + (\mu_(0) i)/(2\pi R_(2) )`

`B =( 4\pi *10^(-7) *5)/(2\pi ) [(1)/(0.03) + (1)/(0.05) ]`

`B = 5.33*10^(-5) T`

Therefore, the magnitude of magnetic field at given point =
`5.33*10^(-5) T`