Chapter 20, Problem 33P

To determine

The magnetic field at point P due to both the wires.

Answer to Problem 33P

The net magnetic field at point P due to both the wires is $1.1\times {10}^{-4} \text{T}$

Explanation of Solution

Given:

The given figure is shown below.

  

Both thin parallel wires are 13 cm apart carrying 25 A currents in the same direction. Distance from both the wires to point P is 12 cm and 5 cm.

Formula used:

The magnetic field due to the current in the wire is given by

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

Consider the figure shown below.

  

The magnetic field from the closer wire is

  $\begin{array}{l}{B}_1=\frac{{\mu }_0{I}_1}{2\pi r}=\frac{{\text{(4π×10}}^{\text{-7}}\text{Tm/A)(25A)}}{\text{2π(0}\text{.05m)}}\\ B_1=1\times {10}^{-4}\text{T}\end{array}$

And, the magnetic field from the farther wire is

  $\begin{array}{l}{B}_2=\frac{{\mu }_0{I}_2}{2\pi r}=\frac{{\text{(4π×10}}^{\text{-7}}\text{Tm/A)(25A)}}{\text{2π(0}\text{.12m)}}\\ B_2=4.167\times {10}^{-5}\text{T}\end{array}$

Now, the angle of the line joining both the wires and the farther wire with the point P is

  ${\theta }_1={\tan }^{-1}\frac{5}{12}=22.6°$

Using angle and the individual magnetic field, the magnetic field in the horizontal and vertical direction can be calculated such as

  $\begin{array}{l}{B}_{net x}=B_1\cos ({90}^{\circ }-{\theta }_1)-B_2\cos {\theta }_2\\ B_{net x}=(1\times {10}^{-4}T)\cos {67.4}^0-(4.167\times {10}^{-5}T)\cos {22.6}^0\\ B_{net x}=-4.1\times {10}^{-8}T\approx 0T\end{array}$

  $\begin{array}{l}{B}_{net y}=B_1\sin (90°-{\theta }_1)+B_2\sin {\theta }_1\\ B_{net y}=(1\times {10}^{-4})\sin 67.4°+(4.167\times {10}^{-5})\sin 22.6°\\ B_{net y}=1.1\times {10}^{-4}\text{T}\end{array}$

Thus, the net magnetic field on the point P will be

  $B_{net}=1.1\times {10}^{-4}\text{T} \text{at }90°$

Conclusion:

The net magnetic field at point P due to both the wires is $1.1\times {10}^{-4} \text{T}$