Question

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the wire?

Answer 1

The electric current in the wire is 0.8 A

Step-by-step explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

`B=(2\pi*a )/(u*I)`

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

`I=(2\pi*a*B )/(u)` Formula(2)

`B=0.2*10^(-5)  T = 0.2*10^(-5) (weber)/(m^(2) )`

a =8cm=0.08m

`u=4*\pi *10^(-7) (Weber)/(A*m)`

We replace the known information in the formula (2)

`I=(2\pi*0.08*0.2*10^(-5)  )/(4\pi *10x^(-7) )`

I=0.8 A

Answer: The electric current in the wire is 0.8 A