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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 Fig. 6.06. (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. |
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Answer» Solution :(a) The magnetic field due to straight wire has magnitude `B_(1)=((mu_(0)I))/(2pir)`, at a distance R from the wire and according to right hand rule `B_(1)` points directly into the page within the loop. The magnetic flux `phi_(2)` linking the loop is `phi_(12) = INT vceB_(1).dvceA_(2)`, where `dA_(2)` is the area of an elementary strip as shown in Fig. 6.07 having a value `dA_(2) = a dr` and r ranges from `r_(min) = r to r_(max) = (x + a)`. Therefore, `phi_(12) = int_(x)^(x+a)((mu_(0)I)/(2pir)).adr = (mu_(0)Ia)/(2pi)int_(x)^(x+z)(dr)/r` `=(mu_(0)Ia)/(2pi) In ((x+a)/x)=(mu_(0)a)/(2pi) In (1+a/x)` Therefore, the MUTUAL inductance `M=M_(12)=phi_(12)/I=(mu_(0)a)/(2pi)In (1+a/x)` (b) The square loop is moving in a NON uniform magnetic field. The magnetic flux linked with the loop at any instant is `phi_(B)= (mu_(0)Ia)/(2pi)In (1+a/x)` `therefore` Induced emf `VAREPSILON=-(dphi_(B))/dt=-(dphi_(B))/dx.0dx/dt=-v(dphi_(B))/dx=-d/dx[(mu_(0)Ia)/(2pi).In(1+a/x)]` `=-v(mu_(0)Ia)/(2pi).((-a//x^(2)))/((1+x/a))mu_(0)/(2pi).(a^(2)v)/(x(x+4)).I` In the present problem `a = 0.1 m, x = 0.2 m, v = 10 ms^(-1), I = 50 A` `therefore |varepsilon|=(4pixx10^(-7))/(2pi).((0.1)^(2)(10))/((0.2)(0.2+0.1))xx50 =1.7xx10^(-5)V.` |
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