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Two parallel long straight conductors lie on a smooth plane surface. Two other parallel conductors rest on them at right angle so as to form a square of side a. A uniform magnetic field B exists at right angle to the plane containing the conductors. Now, conductors starts moving outward witha constant velocity `v_0` at `t=0`. Then, induced current in the loop at any time is `(lamda` is resistance per unit length of the conductors)A. `(aBv_0)/(lamda(a+v_0t))`B. `(aBv_0)/(2lamda)`C. `(Bv_0)/lamda`D. `(Bv_0)/(2lamda)` |
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Answer» Correct Answer - C At `t=t` side of square, `l=(a+2v_0t)` Area, `S=l^2=(a+2v_0t)^2` `phi=BS=B(a+2v_0t)^2` `e=(dphi)/(dt)=4Bv_0(a+2v_0t)` `R=lamda[4l]=4lamda(a+2v_0t)` `:. i=e/r=(Bv_0)/R` |
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