Chapter 20, Problem 43P

To determine

The magnitude of the magnetic field at the midpoint between both the wires.

Answer to Problem 43P

The magnitude of the net magnetic field at midpoint between both the wires is $4.12\times {10}^{-5} \text{T}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The current in the top wire is 20 A and the current in the bottom one is 5A at a distance of $20\text{ cm}$

Formula used:

The magnetic field is calculated as

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

The orientation of both the wires is such that the top wire carries current into the plane of the paper and the bottom one carries it towards the right. Thus, according to the right-hand rule, curled fingers indicate the direction of magnetic field. But, their magnetic field is not in the same dimension that is the magnetic field in the top wire is towards the right and in the bottom one is out of the plane of the paper. Thus, the net magnetic field will be the hypotenuse of the right angle they make.

  $\begin{array}{l}{B}_{net}=\sqrt{{\left(\frac{{\mu }_0{I}_{top}}{2\pi r_{top}}\right)}^2-{\left(\frac{{\mu }_0{I}_{bottom}}{2\pi r_{bottom}}\right)}^2}\\ B_{net}=\frac{{\mu }_0}{2\pi r}\sqrt{I_{top}^2+I_{bottom}^2}\\ B_{net}=\frac{{\text{(4π×10}}^{\text{-7}}\text{)}}{\text{2π(0}\text{.100)}}\sqrt{{\text{(20A)}}^{\text{2}}{\text{+(5A)}}^{\text{2}}}\\ B_{net}=4.12\times {10}^{-5}\text{T}\end{array}$

Conclusion:

The magnitude of the net magnetic field at the midpoint between both the wires is $4.12\times {10}^{-5} \text{T}$ .