A closely wound, circular coil with a diameter of 4.60 cm has 700 turns and carries a current of 0.600 A.

What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 9.50 cm from its center?

I got 1.15 x 10^-2 T for the magnitude of the magnetic field at the center of the coil, but I don't know how to calculate this question

Question

A closely wound, circular coil with a diameter of 4.60 cm has 700 turns and carries a current of 0.600 A.

What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 9.50 cm from its center?

I got 1.15 x 10^-2 T for the magnitude of the magnetic field at the center of the coil, but I don't know how to calculate this question

anonymous · Answer

$$B=(munIR^2)/2\int\limits_{x1}^{x2}dx'/['(x-x)^2+R^2]^(3/2)$$

anonymous · Answer

Still confused on this

anonymous · Answer

http://www.phys.uri.edu/~gerhard/PHY204/tsl215.pdf try this:)

anonymous · Answer

so would my radius be 9.5 cm or half of that?

anonymous · Answer

radius will be 2.30cm and x1=0 ;x2=9.5cm

anonymous · Answer

and x would equal... ?

anonymous · Answer

x=9.5 and x' will be variable.......k

anonymous · Answer

so x and x2 are both 9.5. or is x the variable k after the integral?

anonymous · Answer

nope it is constant

anonymous · Answer

sorry, but I just don't understand. after the integral, is x still 9.5 and x2 9.5?