91Ó°ÊÓ

a. Current is$i=10A$.

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

Formulae:

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

$B=\frac{{\mathrm{μ}}_{0}iâˆ\dots }{4\mathrm{Ï€}R}$

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 $=\mathrm{Ï€}rad$.

$B_a=\frac{{\mathrm{μ}}_{0}iâˆ\dots }{4\mathrm{Ï€}R}+\frac{{\mathrm{μ}}_{0}i}{4\mathrm{Ï€}R}+\frac{{\mathrm{μ}}_{0}i}{4\mathrm{Ï€}R}$

$B_a=\frac{{\mathrm{μ}}_{0}i\mathrm{π}}{4\mathrm{π}R}+\frac{2{\mathrm{μ}}_{0}i}{4\mathrm{π}R}$

$B_a=\frac{{\mathrm{μ}}_{0}i}{4R}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

$B_a=\frac{{\mathrm{μ}}_{0}i}{R}\left(\frac{1}{4}+\frac{1}{2\mathrm{π}}\right)$

Substitute the values and solve as:

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

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

$B_a=1.0×{10}^{-3}\text{T}$

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 $B_b=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}+\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

$B_b=2\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

$B_b=\frac{{\mathrm{μ}}_{0}i}{\mathrm{π}R}$

Substitute the values and solve as:

$B_b=\frac{4\mathrm{π}×{10}^{-7}\left(10\right)}{\mathrm{π}\left(5.0×{10}^{-3}\right)}$

$B_b=8.0×{10}^{-4}\text{T}$

Direction of the magnetic field at b:

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