Q 16. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v= 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.


 

(a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).
Magnetic flux associated with the element dy,d=BdA
Where, dA = Area of element dy = a dy
B = Magnetic field at distance y
=μ0I2πy
I = Current in the win

μ0= Permeability of free space =4π×107=μ0Ia2πdyyϕ=μ0Ia2πdyy

v tends from x to a + x.

ϕ=μ0Ia2πxa+ydyy=μ0Ia2π[logey]xa+x=μ0Ia2πloge(a+xx)

For mutual inductance M, the flux is given as:

ϕ=MIMI=μ0Ia2πloge(ax+1)M=μ0a2πloge(ax+1)

(b) Emfinduced in the loop, e = B'av

=(μ0I2πx)av Given, I=50Ax=0.2ma=0.1mv=10m/se=4π×107×50×0.1×102π×0.2e=5×105V