91Ó°ÊÓ

We need to find the strength and direction of the magnetic field at the center of the loop.

The wire and the loop generate magnetic fields at the center of the loop. Hence, the net magnetic field is the sum of the two magnetic fields at the center. In the case of a current-carrying wire, it produces a magnetic field at a distance given by equation

$B_{wire}=\frac{{\mathrm{μ}}_o}{2\mathrm{Ï€°ù}}\frac{I}{r}\left(1\right)$

With current $I$and radius $R$for one loop , its magnetic field is given by equation

$B_{loop}=\frac{{\mathrm{μ}}_{o}I}{2\mathrm{π}R}\left(2\right)$

The distance is the same $r=R$,So, at the center megnatic field is

$B_{center}=\frac{{\mathrm{μ}}_o}{2\mathrm{Ï€}}\frac{I}{R}+\frac{{\mathrm{μ}}_{o}I}{2\mathrm{Ï€¸é}}=\frac{{\mathrm{μ}}_o}{2}\frac{I}{R}\left(\frac{1}{\mathrm{Ï€}}+1\right)\left(3\right)$

Putting values for $I,Rand{\mathrm{μ}}_o$in equation $\left(3\right)$to get $B_{center}$

$$\begin{gathered} B_{center}=\frac{{\mathrm{μ}}_o}{2}\frac{I}{R}\left(\frac{1}{\mathrm{π}}+1\right) \\ =\frac{\left(4\mathrm{π}×{10}^{-7}\mathrm{T}\right)×\left(m/A\right)\left(5A\right)}{2\left(0.01m\right)}\left(\frac{1}{\mathrm{π}}+1\right) \\ =4.1×{10}^{-4}T \end{gathered}$$

If after applying the right-hand rule the direction of the magnetic field is out of the page.