The figure shows two parallel loops of wire having a common axis. The smaller lo

Question

The figure shows two parallel loops of wire having a common axis. The smaller loop (radius r) is above the larger loop (radius R) by a distance x>>R. Consequently, the magnetic field due to the counterclockwise current i in the larger loop is nearly constant throughout the smaller loop. Suppose that x is increasing at the constant rate dx/dt = v. (a) Find an expression for the magnetic flux through the area of the smaller loop as a function of x. In the smaller loop, find (b) an expression for the magnitude of the induced emf and (c) the direction of the induced current.

Explanation / Answer

If we take an small element on loop then it will provide us magnetic field at an angle on taking component we find that Cos componet get cancel.
Therefore
dBX = dBSin(theta)
dBx = (uoIR / (4Pi*r3)) dl
Where r = (X2 + R2)1/2 where X is the distance between the two loops and R is the radius of the loop.
dBx = (uo(IR) / (4Pi)(X2 + R2)3/2)dl
On integrating
B = uoIR2 / 2*(X2+R2)3/2
Magnetic flux = Magnetic field*Area
   = B*(Pi*r2)
(b) Emf = Change in magnetic flux / time
   = d(Magnetic flux)/dt = (-3/2)(Pi*uo*I*R2*r2)X/(X2 +R2)1/2
(c) Direction of induced current will oppose the change in magentic flux therefore direction will be counterclockwise.

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