Why does the magnetic field formula use μ/2π instead of μ/4π?

[Julia Coggins]

From "Physics for Scientists and Engineers" by Randall D. Knight page 1006. A 1 m long, 1.0 mm diameter nichrome heater wire is connected to a 12 V battery. Find the magnetic field strength 1 cm away from the wire. This problem is solved in the book, but they calculate the magnetic field with: μ/2π instead of μ/4π that is in the Biot-Savart Law. Everything else in my calculations was correct, so I'm wondering why they used this? The full equation they used is: B=μ/2π(I/d) related from the Biot Savart Law for a current carrying wire B ={μ/4π}{IΔs×r/r^2} the 'r' in the numerator is the direction of r not the magnitude.

 

[robphy]

Did you see the preceding example (continued onto the top of page 1006)
where they used Biot-Savart to obtain the equation for B_wire?

 

[Julia Coggins]

Reading it now. I don't understand how they evaluated that integral?

 

[robphy]

I don't understand how they evaluated that integral?

https://www.wolframalpha.com/input/?i=integral(1/sqrt(x^2+d^2)^3,x) (Trig substitution... see step-by-step)
The text gave you the integral to do, then they did the integral (with limits).
Plug in the limits and use l'Hopital's rule... you get "1 - (-1)" = 2

 

[Julia Coggins]

Unfortunately I don't have wolfram alpha pro, I do need to brush up on my calc though. Isn't l'hospital's rules dealing with indefinite integrals like this one?