Question

What are the magnetic field strength and direction at points a to c in FIGURE EX29.14?

Short answer

The magnetic field strength is

$$\begin{gathered} B_a=66.7\times {10}^{-6}T \\ {B}_b=200\times {10}^{-6}T \\ {B}_c=66.7\times {10}^{-6}T \end{gathered}$$

Step-by-step solution

Given information

We have given,

The magnetic field strength and direction at points a to c in the figure as shown above.

current in the wires are $10\mathrm{A}.$

We have to find the magnetic field at a ,b, c.

Simplify

The magnetic field due to current carrying wire is given by

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

The upper wire carrying current to the right direction while the bottom wire has current to the left and both of them have same magnitude of the current .

Since point a is nearer to upper wire and far away to lower wire, then magnetic field will be more at a due to upper wire.

The magnetic field at point a,

localid="1649343205269" $$\begin{gathered} B_a=\frac{{\mu }_{o}I}{2\mathrm{\pi }}\left(\frac{1}{r_{up}}-\frac{1}{r_{down}}\right) \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)}{2\mathrm{\pi }}\left(\frac{1}{0.02m}-\frac{1}{0.06m}\right) \\ =66.7\times {10}^{-6}T \end{gathered}$$

Here positive sign indicates the magnetic field is out of the paper.

Since at point b both the wires generates same directional magnetic field then here we need to add the magnetic field due to both the wire and the distances from both wires are same then magnetic field will be same.

localid="1649343218625" $$\begin{gathered} B_b=\frac{{\mu }_{o}I}{2\mathrm{\pi }}\left(\frac{1}{r_{up}}+\frac{1}{r_{down}}\right) \\ =2\times \left(\frac{4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}}{2\mathrm{\pi }}\right)\times \frac{1}{0.02} \\ =20\times {10}^{-5}\mathrm{T} \end{gathered}$$

Since at c the magnetic field will be in opposite direction. then magnetic field will be

localid="1649343229931" $$\begin{gathered} B_c=\frac{{\mu }_{o}I}{2\mathrm{\pi }}\left(\frac{1}{r_{up}}-\frac{1}{r_{down}}\right) \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)}{2\mathrm{\pi }}\left(\frac{1}{0.02m}-\frac{1}{0.06m}\right) \\ =66.7\times {10}^{-6}T \end{gathered}$$