your question is missing the figure which I have attached below;
Answer:
[tex]B=198.33*10^{-7}[/tex]Tesla.
Explanation:
Using Biot-Savart law
[tex]dB=[/tex]μ[tex]\frac{(I_{ds}r) }{4\pi\ r^{3} }[/tex]
[tex]B=\frac{u}{4\pi}\int\limits^a_b {\frac{I_{ds}r }{r^{3} } }[/tex]
as we know from the figure that cure 1/4 of the circle so integrating the above equation from 0 to 2pi and then dividing it by 4 we get;
[tex]B=\frac{u}{4\pi}\int\limits^{2\pi}_{0} {\frac{Irsin(90) }{r^{3} } }\,\ ds[/tex]
[tex]B=\frac{uI}{8r}[/tex]
Also we know that magnetic field in the length of a wire is
[tex]B=\frac{uI}{4\pi\ r}[/tex]
in the figure there are two parts as length other than curve part so for both parts the sum of magnetic field will be
[tex]B=\frac{uI}{2\pi\ r}[/tex]
Now the total magnetic field is given as
[tex]B=\frac{uI}{8r}+\frac{uI}{2\pi\ r}[/tex]
[tex]B=\frac{uI}{8\pi\ r}(\pi+4)[/tex]
putting values we get
[tex]u=4\pi*10^{-7} T.m/A[/tex]
[tex]I=15A[/tex]
[tex]r=27cm=0.27m[/tex]
so
[tex]B=\frac{4\pi*10^{-7}*15}{8*3.14*.27}(3.14+4)[/tex]
solving above equation we get
[tex]B=198.33*10^{-7}[/tex]Tesla.
