A constant current `I` flows through a long straight wire as shown in figure. A square loop starts moving towards righ with as constant speed `v`.

a Find induced emf produced i the loop as a function `x`
b. If total resistance of the loop is `R`, then find induced current in the loop,

Explanation

Method 1
a. Using the result of magnetic flux obtained in example 1, we have
`phi_B=(mu_0IC)/(2pi)In((a+b)/c)`
Here `a=x,b=c=a`
substituting the values, we get
`phi_B-(mu_0Ia)/(2pi) ln ((x+a)/x)`
`=(mu_0Ia)/(2pi) ln (1+a/x)`
Now, `|e|=|(dphi_B)/(dt)|=(mu_0Ia)/(2pi) (x/(x+a))(a/x^2)(dx)/(dt)`
Putting `(dx)/(dt)=v`, we have
`|e|=(mu_0Ia^2)/(2pix(x+a)^v`
b. induced curent
`I_(in) =(|e|)/R=(mu_0Ia^2v)/(2piRx(x+a))`
Method 2

`e_1=B_1vl=mu_0/(2pi)I/xva`
`e_2=B_2vl=mu_0/(2pi) I/(x+a) va`
`e_(1) gt e_(2)`
`e_("net")=e_1-e_2=(mu_0Iva)/(2pi) (1/x-1/(x+a))`
`=(mu_0Iva^2)/(2pix(x+a))`
This is the name result as was obtained in method 1.