Answer:
13.2 A
Explanation:
Earth's magnetic field B= 5.5×10^{-5} T
the distance r= 4.8 m
the permeability of free space μ_o= [tex]4\pi\times10^{-7}T.m/A[/tex]
according to Biot Savart law
[tex]B= \frac{\mu_0 I}{2\pi r}[/tex]
therefore, current in the table is
[tex]I= \frac{B(2\pi r)}{\mu_0}[/tex]
[tex]= \frac{5.5\times10^{-5}\times2\pi\times4.8}{4\pi\times10^{-7}}[/tex]
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%)
= [tex]\frac{1}{100}\times1320[/tex]
= 13.2 A
