Question

The magnetic field at the center of $1.0\mathrm{cm}$diameter loop is $2.5mT$.

a. What is the current in the loop?

b. A long straight wire carries the same current you found in part a. At what distance from the wire is the magnetic field$2.5mT$?

Short answer

(a) $I=20A$.

(b)$r=1.6mm$

Step-by-step solution

Part (a) Step 1 : Given Information 

We have given,

The magnetic field at the center of a$1.0cm$diameter loop is $2.5mT$.

We have to find the current in the loop.

Simplify

Each segment on the loop applies a magnetic field, so the total magnetic field exerted by one loop at any point is given by equation

$B_{loop}=\frac{{\mu }_{o}I}{2R}$

Our target is to find the current, so we rearrange equation and to be in the form

$I=\frac{2R{B}_{loop}}{{\mu }_o}$

Now we can plug the values into equation to get the current of the loop

$$\begin{gathered} I=\frac{2R{B}_{loop}}{{\mu }_o} \\ =\frac{2\left(0.005m\right)\left(2.5\times {10}^{-3}T\right)}{4\pi \times {10}^{-7}T·m/A} \\ =20A \end{gathered}$$

Part (b) Step 1: Given information

We have given,

The magnetic field at the center$1.0\mathrm{cm}$is$2.5\mathrm{mT}.$

We have to find the distance from the wire.

Simplify

At any point where the magnetic field due to the segment of current-carrying wire is given by equation

$B_{wire}=\frac{{\mu }_o}{2\pi }\frac{I}{r}$

We use the value of the current as the same for the loop $I=20A$
.Now , plug the values

$$\begin{gathered} r=\frac{{\mu }_o}{2\pi }\frac{I}{B_{wire}} \\ =\frac{\left(4\pi \times {10}^{-7}T·m/A\right)\left(20A\right)}{2\pi \left(2.5\times {10}^{-3}T\right)} \\ =1.6\times {10}^{-3}m \\ =1.6mm \end{gathered}$$