induced emf

Question

A infinite long wire carrying a current I=ie^-t . Where t is the time. At a distace d from the wire and centre of rectangle's centre , a conducting rectangle (a x b) frame is kept . Find the induced emf across the rectangular loop

Answer 1

David is absolutely right about the geometry, but I'll take a stab at it with a (what I consider reasonable) assumption.  It sounds in the description of the the distance d is that it is the radial distance from the wire to the center of the rectangle.

 

The one assumption I will make is that one pair of sides of the rectangle is parallel to the wire, and the other pair is in the radial direction.  There is no indication which side is which, but that only makes a minor change to the overall solution.  

 

I will take the sides of length b to be in the radial direction, and the sides of length a to be parallel to the wire.

 

The induced emf is related, by Faraday's law, to the time rate of change (in other words, the time derivative) of flux, as:

 

emf = -N(dΦ_b/dt)

 

where 

 

N = number of turns of coil = 1 in this case

Φ_B = magnetic flux

 

In turn, 

 

Φ_B = BAcos(θ)

 

where

 

B = total magnetic field in loop

A = area of loop

θ = angle between the "area vector" of the loop and the magnetic field

 

Now, the "area vector" of a loop is perpendicular to the plane of the loop, and -- if the loop is in the configuration I described above -- so is the magnetic field from the wire.  Hence, the magnetic field and the area vector are in the same direction (parallel), and the angle between them is zero (they could also be anti-parallel, but the effective result would be the same).  So, cosθ in the magnetic flux expression is always 1 (cos(0^o)).

 

The area of the loop is just A = ab, as for any rectangle.

 

The total magnetic field through the loop is trickiest to get.  The strength of the field depends on the radial distance away from the wire.  As that distance changes, so does field strength.  So it is not constant across the area of the wire.  It changes continuously with radial distance, and therefore has to be integrated.

 

The expression for the magnetic field as a function of radial distance r from a long, straight wire is given by:

 

B(r) = (μ_oI)/(2πr)

 

If I call the direction parallel to the wire "z," then an infinitesimal area unit for flux to go through in the rectangle .is drdz.  We need to integrate B over the rectangle, and I defined the rectangle above so that the "b" sides were in the radial direction.  If I presume one side of the rectangle in the parallel (z) direction to be at zero, and the other side at "a," then I have to integrate in z from 0 to a, and in r from (d-b/2) to (d+b/2).  The integral for the total magnetic field through the rectangle thus becomes:

 

B_tot = ∫₀^a ∫_d-b/2^d+b/2  (μ_oI/2πr)drdz

 

With respect to this area integral, only the 1/r term inside depends on either r or z.  Everything else can come out.  This means we have:

 

B_tot = (μ_oI/2π)∫₀^a ∫_d-b/2 ^d+b/2  (1/r)drdz = (μ_oIa/2π)[ln((2d+b)/(2d-b))]

 

(I believe this is correct for the integral; if you want to go into more detail about how to get that, just let me know)

 

Then, the total flux becomes:

 

Φ_B = B_totAcos(θ) = (μ_oIa/2π)[ln((2d+b)/(2d-b))](ab)(1) =  (μ_oIa²b/2π)[ln((2d+b)/(2d-b))]

 

Now, back to Faraday's law

 

emf =  -N(dΦ_B/dt)

 

The negative sign is immaterial for our case; it only really comes into play figuring out the direction of induced current, in most cases.  And there is only one loop, so N = 1.

 

The only part of the magnetic flux expression that depends on time is the current, do dΦ_B/dt becomes

 

dΦ_B/dt = (μ_oa²b/2π)[ln((2d+b)/(2d-b))](dI/dt), with dI/dt = (d/dt)(ie^-t)

 

I take i to be a constant coefficient, since no time dependence is given.  In that case, 

 

(d/dt)(ie^-t) = -ie^-t

 

So 

 

dΦ_B/dt = -(μ_oa²b/2π)[ln((2d+b)/(2d-b))](ie^-t)  [NOTE: the minus sign only indicates decrease, so the magnitude at any given instant is just this without the minus sign in front; that is what I will state as the solution; the fact that it is decreasing will affect the direction of induced current, but not the magnitude of the induced emf, which it seems the problem is asking for]

 

So:

 

emf(t) = ((μ_oa²bie^-t)/2π)[ln((2d+b)/(2d-b))]

 

No absolute guarantees on all the algebra, but the process should be correct.  If the a sides of the rectangle are in the radial direction and the b sides in the tangential direction, just switch a for b in the above expression.  If the rectangle is oriented so that neither pair of sides is in the radial direction, then the math gets more involved... though it adds complication without adding anything to the physical understanding (so I am guessing that is not the case here).

 

Hope this helps!  Let me know if you have any questions, or if my assumption(s) is (are) wrong.