Question

What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in air? Give your answer in units of tesla.

Answer 1

The magnetic flux density is
`2.11*10^(-6)\ T`

Step-by-step explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

`B =(\mu_(0)I)/(2r)`

Where,

r = radius

I = current

Put the value into the formula

`B =(4\pi*10^(-7)*3.8)/(2*\pi*0.36)`

`B=2.11*10^(-6)\ T`

Hence, The magnetic flux density is
`2.11*10^(-6)\ T`