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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. |
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Answer» Solution :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` |
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