Question

An experiment on the Earth's magnetic field (5.5 10-5 T) is being carried out 4.8 m from an electric cable. What is the maximum allowable current in the cable if the experiment is to be accurate to 1.0% ?

Answer 1

13.2 A

Step-by-step explanation:

Earth's magnetic field B= 5.5×10^{-5} T

the distance r= 4.8 m

the permeability of free space μ_o=
`4\pi*10^(-7)T.m/A`

according to Biot Savart law

`B= (\mu_0 I)/(2\pi r)`

therefore, current in the table is

`I= (B(2\pi r))/(\mu_0)`

`= (5.5*10^(-5)*2\pi*4.8)/(4\pi*10^(-7))`

solving we get I= 1320 A

if the experiment is to be accurate to 1.0%

then the allowed value of current in the cable

I'= I(1%)

=
`(1)/(100)*1320`

= 13.2 A