Hello!
a)
To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.
Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]
B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)
i = Current in wire ('I' A)
dl = differential length element
r = distance from wire to point P ('R' m, remains constant!)
We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]
First, let's mind the cross-product.
The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.
Therefore, we can disregard the cross-product since sin(90) = 1.
Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]
We have a semi-circle, so we are integrating from 0 to π radians.
[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]
Simplifying to make the integral easier, we can take constants out of the integral.
[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]
Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]
b)
Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.
