In Figure, a current $\mathbf{i}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{}\mathbf{A}\mathbf{}$is set up in a long hairpin conductor formed by bending a wire into a semicircle of radius$\mathbf{R}\mathbf{}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mm}$. Point bis midway between the straight sections and so distant from the semicircle that each straight section can be approximated as being an Infinite wire. (a)What are the magnitude and (b) What is the direction (into or out of the page) of $\overrightarrow{\mathbf{B}}$at aand (c) What are the magnitude and (d) What is the direction $\overrightarrow{\mathbf{B}}$ of at b?

a. Current is$i=10A$.

b. Radius of semicircle is:$\begin{array}{rcl}R& =& 5.0\text{mm}\\ & =& 5.0\times {10}^{-3}\text{m}\\ & & \end{array}$

Formulae:

$B=\frac{{\mu }_{0}i}{2\pi R}$

$B=\frac{{\mu }_{0}i\varnothing }{4\pi R}$

Magnitude of the magnetic field at a:

At point a, we can write the total magnetic field due to a semicircular arc and two semi-infinite straight wires. Here angle $=\pi rad$.

$B_a=\frac{{\mu }_{0}i\varnothing }{4\pi R}+\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{4\pi R}$

$B_a=\frac{{\mu }_{0}i\pi }{4\pi R}+\frac{2{\mu }_{0}i}{4\pi R}$

$B_a=\frac{{\mu }_{0}i}{4R}+\frac{{\mu }_{0}i}{2\pi R}$

$B_a=\frac{{\mu }_{0}i}{R}\left(\frac{1}{4}+\frac{1}{2\pi }\right)$

Substitute the values and solve as:

$B_a=\frac{4\pi \times {10}^{-7}\left(10\right)}{\left(5.0\times {10}^{-3}\right)}\left(\frac{1}{4}+\frac{1}{2\pi }\right)$

$B_a=\frac{4\left(3.14\right)\times {10}^{-7}\left(10\right)}{\left(5.0\times {10}^{-3}\right)}\left(\frac{1}{4}+\frac{1}{\left(2\times 3.14\right)}\right)$

$B_a=1.0\times {10}^{-3}\text{T}$

Direction of the magnetic field at a:

Using the right hand rule the direction of magnetic field is out of page.

Magnitude of the magnetic field at b:

At point b, the magnetic field would be due to two infinite wires so we can write $B_b=\frac{{\mu }_{0}i}{2\pi R}+\frac{{\mu }_{0}i}{2\pi R}$

$B_b=2\frac{{\mu }_{0}i}{2\pi R}$

$B_b=\frac{{\mu }_{0}i}{\pi R}$

Substitute the values and solve as:

$B_b=\frac{4\pi \times {10}^{-7}\left(10\right)}{\pi \left(5.0\times {10}^{-3}\right)}$

$B_b=8.0\times {10}^{-4}\text{T}$

Direction of the magnetic field at b:

From the right hand rule the direction of magnetic field is out of page.