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Now assume that the straight wire carries a current of 50A and the loop is moved to the right with a constant velocity, `upsilon=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» Given current `I=50A, "Velocity" V=10m//s, x=0.2m and a=0.1m` Induce emf e `=(-d theta)/(dt)=(-d)/(dt)[I(mu_(0)a)/(2pi)log_(e)((a+x)/(x))]` `e=(-mu_(0)aI)/(2pi).d/(dt)(log(x+a)-logx)` `=(-mu_(0)aI)/(2pi)((1)/((x+a))d/dt(x+a)-1/x.d/dt(x))=(-mu_(0)aI)/(2pi)((V)/(x+a)-(V)/(x))` `=(-mu_(0)aI)/(2pi).(a^(2).IV)/(x(a+x))Rightarrow e=(4pixx10^(-7))/(2pi)xx((0.1)^(2)xx(50)xx(10))/(0.2xx(0.1+0.2))=(2xx10^(-7)xx50xx10^(-2)xx10)/(0.2xx0.3)` `e=1.67xx10^(-5)V` |
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