a. Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in the figure.

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=10m//s`. Calculate the induced emf in the loop at the instant when `x=0.2m`. Take `a=0.1m` and assume that the loop has a large resistance.

Explanation

a. Consider a small rectangular element of the square, of thickness dt at a distance .t. from the conductor carrying a current I.
Magnetic field experienced by this element `=B=(mu_(0).I)/(2pit)`
Area of the element = `dA=a.dt`
`therefore` Flux, `dphi_(B)=B.dA=(mu_(0)Ia)/(2pit)dt`
Hence net flux, `phi_(B)=underset(t=x)overset(t=a+x)int(mu_(0)Ia)/(2pit)dt=(mu_(0)Ia)/(2pi)log_(e)(1+(a)/(x))`
But `phi_(B)=MI`, M - Mutual inductance
`therefore M=(mu_(0)a)/(2pi)log_(e)(1+(a)/(x))`
b. `varepsilon=(mu_(0)Ia^(2)v)/(2pix(a+x))=(4pixx10^(-7)xx50xx(0.1)^(2)xx10)/(2pixx0.2(0.1+0.2))=1.66xx10^(-5)=1.7xx10^(-5)V`