Answer:
[tex]B=331.32*10^{-7}T[/tex]
Explanation:
Given data
The current in the loop I=0.51 A
The radius of loop r=14.5m=0.145 m
Number of turns of coil n=15 turns
To find
Magnetic field B
Solution
Each segment on loop applies a magnetic field,so total magnetic field exerted by loop at any point is given by:
[tex]B=\frac{u_{o}IR^2j}{2\pi (y^2+R^{2} )^{\frac{3}{2} }}[/tex]
At the center of loop y=0 so the magnitude of B of n loops of a coil
So
[tex]B=\frac{u_{o}nI}{2R}[/tex]
Now plug the values to get the magnitude of B
So
[tex]B=\frac{(4\pi *10^{-7}T.m/A)*(15)*(0.51A)}{2(0.145m)} \\B=331.32*10^{-7}T[/tex]
