Answer:
Current, I = 5.3 A
Explanation:
It is given that,
Radius of circular loop, r = 3 mm = 0.003 m
Magnetic field strength, B = 1.1 mT = 0.0011 T
We need to find the magnitude of the current that flows through the wire. The magnetic field for a current carrying wire is given by :
[tex]B=\dfrac{\mu_o I}{2r}[/tex]
[tex]I=\dfrac{2Br}{\mu_o}[/tex]
[tex]I=\dfrac{2\times 0.0011\ T\times 0.003\ m}{4\pi\times 10^{-7}}[/tex]
I = 5.25 A
or
I = 5.3 A
So, the magnitude of the current that flows through the wire is 5.3 A. Hence, this is the required solution.
