Answer:
The net magnitude of magnetic field at the common center is
[tex]3.729 \times 10^{-5}[/tex] T
Explanation:
Given:
Radius of circular loop [tex]r = 5 \times 10^{-2}[/tex] m
Current [tex]I = 2.10[/tex] A
Number of turn [tex]N = 1[/tex]
Magnitude of magnetic field at the common center is given by,
[tex]B_{1} = \frac{\mu_{o} NI }{2r}[/tex]
[tex]B _{1} = \frac{4\pi \times 10^{-7} \times 1 \times 2.10 }{2 \times 5 \times 10^{-2} }[/tex]
[tex]B_{1} = 2.637 \times 10^{-5}[/tex] T
Now, both loops having same magnetic field but perpendicular to each other.
[tex]B _{net} = \sqrt{B_{1}^{2} + B _{1}^{2} }[/tex]
[tex]B _{net} = \sqrt{2} B_{1}[/tex]
[tex]B_{net} = \sqrt{2} \times 2.637 \times 10^{-5}[/tex]
[tex]B_{net} = 3.729 \times 10^{-5}[/tex] T
Therefore, the net magnitude of magnetic field at the common center is [tex]3.729 \times 10^{-5}[/tex] T
