Question

A long straight wire carries a current I1 = 15A and has a length 1.1m. (a) What is the magnetic field created by this wire a distance of 0.1m away? (b) A second wire has the same length and carries a current I2 = 5A in the same direction as the first wire. If this wire is placed 0.1m away from the first, what is the force it experiences? In what direction is the force on the second wire?

Answer 1

a) 2.95*10^{-5}T

b) 1.65*10^{-4}N

Explanation:

(a) The magnetic field produce by a finite wire, in a poin perpendicular to the center of the wire, is given by

`B=(\mu_0 I)/(4\pi R)(2)\int_(0)^(\phi)cos\phi d\phi` (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:

`tan\phi=(0.55)/(0.1)=5.5\\\phi=tan^(-1)(5.5)=79.69\°`

Hence, by replacing in (1) we obtain:

`B=((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`

(b) The magnetic force of two wires is given by:

`F=(\mu_0 I_1 I_2 L)/(2\pi r)=((4\pi*10^(-7)Tm/A)(15A)(5A)(1.1m))/(2\pi(0.1m))=1.65*10^(-4)N`

hope this helps!!