Question

A wire placed along north south direction carries a current of 5 A from South to North. Find the magnetic field due to a 1 cm piece of wire at a point 200 cm North East from the piece

Answer 1

see diagram.

Using Biot-Savart's law:

$\vec{dB}=\frac{\mu_0i}{4\pi}\frac{dl \times \hat{r}}{r^2}\\\\\vec{dB}\ has\ same\ direction\ \perp to\ diagram\ for\ all \ length\ of\ wire.\\\\dB=\frac{\mu_0i\ dl\ sin\theta}{4\pi r^2}.\ \ \ ---(1)\\\\l=100\sqrt2-d\ cot\theta,\ \ dl=d\ cosec^2\theta d\theta\\r=d\ cosec\theta\\\\dB=\frac{\mu_0 i}{4 \pi d}sin\theta d\theta,\ \ \ ----(2)\\\\B=\frac{\mu_0 i}{4 \pi d}(Cos\theta_1 - Cos\theta_2)\ \ \ \ ----(3)\\\\d\theta=\frac{dl\ sin\theta2}{r}=\frac{1cm*sin\ 45^0}{200cm}$

Here, the length of wire segment is so small compared to the distance of the point, we can use any of the 3 formulas above to find dB or B.

Using (3) is difficult as finding  θ1 and θ2 and their Cosine is tough.  So use either (1) or (2) directly.

Using (1):  

dB = 10⁻⁷ units * 5A * 0.01 m * Sin 45⁰ / (2 m) ²      Tesla
      = 1.77 * 10⁻⁷ Tesla