Answer:
687.5 A
Explanation:
We are given that
[tex]\mu_0=1.25664\times 10^{-6}N/A^2[/tex]
Magnetic field,B=0.11 T
Diameter,d=2.5mm
Radius,r=[tex]\frac{d}{2}=\frac{2.5}{2}mm=1.25\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
We know that current in coil
[tex]I=\frac{2\pi rB}{\mu_0}[/tex]
Substitute the values
[tex]I=\frac{2\pi\times 1.25\times 10^{-3}\times 0.11}{1.25664\times 10^{-6}}[/tex]
[tex]I=687.5 A[/tex]
