Answer:
a) 2.95*10^{-5}T
b) 1.65*10^{-4}N
Step-by-step explanation:
(a) The magnetic field produce by a finite wire, in a poin perpendicular to the center of the wire, is given by
[tex]B=\frac{\mu_0 I}{4\pi R}(2)\int_{0}^{\phi}cos\phi d\phi[/tex] (1)
where I is the current, R is the distance to the wire and phi is the angle between the center of the wire until the top. The number 2 before the integral is because of the symmetry.
In this case we have, by geometry:
[tex]tan\phi=\frac{0.55}{0.1}=5.5\\\phi=tan^{-1}(5.5)=79.69\°[/tex]
Hence, by replacing in (1) we obtain:
[tex]B=\frac{(4\pi*10^{-7}Tm/A)(15A)}{2\pi(0.1m)}\int_{0}^{79.69}cos\phi d\phi\\\\B=3*10^{-5}sin(79.69\°)T=2.95*10^{-5}T[/tex]
(b) The magnetic force of two wires is given by:
[tex]F=\frac{\mu_0 I_1 I_2 L}{2\pi r}=\frac{(4\pi*10^{-7}Tm/A)(15A)(5A)(1.1m)}{2\pi(0.1m)}=1.65*10^{-4}N[/tex]
hope this helps!!
