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Two parallel long straight conductors li...

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 with a 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)`

Text Solution

Verified by Experts

The correct Answer is:
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)/lamda`
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