# Magnetic field due to a current in a straight wire A corner point makes angles θ1 and θ2 with the ends of a straight wire carrying current I. The distance of the point from the wire is a. What is the
magnetic field at the point?

###### Solution.

The magnitude of the field produced at point at distance r by a current-length element i turns out to be

dB = (μ0/4π)*(i*ds*sin(θ)/r2)

where θ is the angle between the directions of vectors ds and r. This equation is known as the law of Biot and Savart.

We can see that

a = r*sin(180-θ) = r*sin(θ) or r = a/sin(θ)     (1)

x = r*cos(180-θ) = –r*cos(θ) or x = -a*cot(θ)     (2)

So,

dx = -a*dcot(θ) = -a*-dθ/sin2(θ) = a*dθ/sin2(θ) (3)

We can find the magnitude of the magnetic field produced at the corner point by integrating dB.

∫dB = I*(μ0/4π)∫dx*sin(θ)/r2 (4)

We can use the equations (1) and (3) for the expression dx/r2 and we get

dx/r2 = dx*sin2(θ)/a2 = a*(dθ/sin2(θ))*sin2(θ)/a2 = dθ/a

or

dx/r2 = dθ/a (5)

Now, we substitute the expression (5) into the equation (4) and we get

∫dB = I*(μ0/(4πa))∫sin(θ)dθ (6)

To find the magnitude of the total magnetic field at the point, we need to integrate from θ=θ1 to θ=θ2.

Thereby,

B = (I*(μ0/(4πa))*(cos(θ1)-cos(θ2)).

Using the right-hand rule we can get that the magnetic field (at the point) is directed out of the page, as indicated by the dot.