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A conducting wire XY of mass M and negli...

A conducting wire XY of mass M and negligible resistance slides smothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field `B=B(t)hatk`

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Fig. shows a conducting wire XY of mass m and negligible resistance sliding smoothly over two parallel conducting wires AB and CD. The magneitic field applied is `vec B = B (t) hat k`
(i) At time t, suppose XY is at x = x (t)
Magnetic flux associated with area ACYX
`phi = B (t).l.x.(t)`
emf induced, `e = (d phi)/(dt) = - [(dB)/(dt) lx (t) + B (t) l.v. (t)]`
current induced , `i = (e)/(R )`
Force on XY, `F = B i l = B ((e)/(R )) l` or `m (d^(2) x)/(dt) = (B(t).l)/(R ) [ -(dB)/(dt) lx (t) - B (t).l.upsilon (t)]`
`:.` Acceleration of wire ,`(d^(2) x)/(dt^(2)) = -(l^(2) B)/(mR) (dB)/(dt) x (t) = (l^(2) B^(2))/(mR) ((dx)/(dt))`
(ii) When B is independent of time, it is constant w.r.t. time. Therefore, `(dB)/(dt) = 0`
From (ii), `(d^(2) x)/(dt^(2)) + (l^(2) B^(3))/( m R) (dx)/(dt) = 0` or `(d upsilon)/(dt) + (l^(2) B^(2))/(m R) upsilon = 0`. `:. upsilon = A exp (-(l^(2) B^(2) t)/(m R))`
At `t = 0, upsilon = u_(0) = A` `:. upsilon (t) = u_(0) exp (-(l^(2) b^(2) t)/(mR))`
(iii) Now, `I^(2) R = (B^(2) l^(2) upsilon^(2) (t))/(R^(2)) xx R` using (iii) `I^(2) R = (B^(2) l^(2))/(R ) u_(0)^(2) exp (-(2 l^(2) B^(2) t)/(mR))`
Power lost `= int_(0)^(t) I^(2) R dt = (B^(2) l^(2))/(R ) u_(0)^(2) (mR)/(2 l^(2) B^(2)) [1 - e^(-l^(2) b^(2) t//mR)] = (m)/(2) u_(2) - (m)/(2) upsilon^(2) (t)`
=decrease in KE of wire XY, which was to proved.
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