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In a given circuit inductor of L = 1m H ...

In a given circuit inductor of L = 1m H and resistance `R=1 Omega` are connected in series to ends of two parallel conducting rods as shown. Now a rod of length 10 cm is moved with constant velocity of 1cm/s in magnetic field B=1T. If rod starts moving at t=0 then current in circuit after 1milisecond is `x=10^(-3)A`. then value of x is : (given `e^(-1)=0.37`)

Text Solution

Verified by Experts

`e=(vec(v)xxvec(B)).vec(dl) =10^(-2) xx1xx10^(-1)`
`epsilon=10^(-3)` volt
`i=(10^(-3))/1 (1-e^(-1))`
`i=10^(-3) (1-0.37)`
i=0.63 mA
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