Question

What are the strength and direction of the magnetic field at the center of the loop in $FIGUREP29.43$?

Short answer

$B_{center}=4.1\times {10}^{-4}T$out of the page.

Step-by-step solution

Step.1  Given information 

We have to find what are the strength and direction of the magnetic field at the center of the loop in $FIGUREP29.43$?

Step.2 solution

The center of the loop experiences both magnetic fields due to the wire and the loop. So, the net magnetic field at the center is the summation of both magnetic field s. For a current-carrying wire, it produces a magnetic field at distance r given by equation $\left(29.7\right)$in the form

localid="1649197843882" $B_{wire}=\frac{{\mu }_o}{2\pi }\frac{I}{r}$ $\left(1\right)$

While for one loop with current I and radius R, its magnetic field is given by equation $\left(29.8\right)$in the form

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

In the given figure both distances is the same $r=R$.So,at the center the magnetic field is

$B_{center}=\frac{{\mu }_o}{2\pi }\frac{I}{R}+\frac{{\mu }_{o}I}{2R}=\frac{{\mu }_o}{2}\frac{I}{R}\left(\frac{1}{\mathrm{\pi }}+1\right)$

Now, we plug the values for I,R and ${\mu }_o$ into equation $\left(3\right)$to get $B_{center}$

localid="1649197616834" $$\begin{gathered} B_{center}=\frac{{\mu }_o}{2}\frac{I}{R}\left(\frac{1}{\pi }+1\right) \\ =\frac{\left(4\pi \times {10}^{-7}T·m/A\right)\left(5A\right)}{2\left(0.01m\right)}\left(1+\frac{1}{\pi }\right) \\ =4.1\times {10}^{-4}T \end{gathered}$$

If we apply the right-hand rule we find that the direction of the magnetic field is Out of the page.