Consider this diagram: Two wires, of different lengths are placed near one anoth
ID: 3280002 • Letter: C
Question
Consider this diagram:
Two wires, of different lengths are placed near one another, and as shown in the diagram. Due to the currents and their magnetic fields... there is a Lorentz force that attracts both wires together within the (yaxis).
However, if I were to somehow apply an external force F to move the wire1(above with current I1) to the (+x)direction:
What is the consequence of the change in area within the magnetic flux() between the two wires?
Also, how would things change considering both wires are finite, and wire2's length is greater(L2> L1)?
Explanation / Answer
As the wires are moving away from each other, lets assume the overlap length be x
then dx/dt = v
now, the area of the associated magnetic flux between the two wires is dependent on the cross section of the wires and distance x
as the cross section is constant, dA/dt = k*dx/dt = kv ( where k is a constant)
so as the magnetic field remains the same at the location of the wire if we consider infinite wires ( as the distance betweens the wires is not changin)
the flux changes as d(phi)/dt = k'v ( where k' is some other constant)
Now if we consider the wires to be finite
then magnetic field at the location of the wire 1 due to wire 2 is given by
B = ki2((L2 - x/2)/sqroot(d^2 + (L - x/2)^2) + (x/2)/sqroot((x/2)^2 + d^2)) /d
where d is distance betweeen the wires and x is the overalp length
so, d(phi)/dt = d(B*A)/dt = d[(k'vt)i2((L2 - x/2)/sqroot(d^2 + (L - x/2)^2) + (x/2)/sqroot((x/2)^2 + d^2)) /d]/dt [ here k' is some constant]
AdB/dt + BdA/dt = k'vt * (i2/d){[sqroot(d^2 + (L - x/2)^2)(-v/2) + v(L2 - x/2)/2sqroot(d^2 + (L - x/2)^2)]/(d^2 + (L - x/2)^2) + [sqroot(d^2 + x^2/4)(v/2) v(x/2)/2sqroot(d^2 + (x/2)^2)]/(d^2 + x^2/4)} + k"v*i2((L2 - x/2)/sqroot(d^2 + (L - x/2)^2) + (x/2)/sqroot((x/2)^2 + d^2)) /d
here k' and k" are some constants
hence we see flux change follows a really complex form for finite wires, but it does decrease with time