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

Answer:

The electric current in the wire is 0.8 A

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:

$B=\frac{2\pi*a }{u*I}$

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):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a  =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A