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A square loop of side 12 cm with its sid...

A square loop of side `12 cm` with its sides parallel to `x` and `y`- axes is moved with a velocity `8 cm//s` along positive `x`-direction in an environment containing magnetic field along `+ve` z-direction. The field has a gradient of `10^(-3) "tesla"//"em"` along `-ve` x-direction (increasing along `-ve` x-axis) and also decreases with time at the rate of `10^(-3) "tesla"//s`. The emf induced in the loop is

Text Solution

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Here, area of loop, `A = (12 xx 10^(-2))^(2) = 144 xx 10^(-4) m^(2)` , velocity, `upsilon = 8 cm s^(-1) = 8 xx 10^(-2) ms^(-1)`
`(db)/(dx) = 10^(-3) T cm^(-1)` `(dB)/(dt) = 10^(-3) T s^(-1), R = 4.5 xx 10^(-3) ohm`
Induced e.m.f. e = ?
Rate of change of magnetic flux due to explicit time variation in `B = A ((dB)/(dt))`
`= (144 xx 10^(-4)) xx 10^(-3) Wb s^(1) = 1.44 xx 10^(-5) Wb s^(-1)`.
Rate of change of magnetic flux due to motion of the loop in non-uniform magnetic field `= A ((dB)/(dx)) ((dx)/(dt))`
`= (144 xx 10^(-4)) xx 10^(-3) xx 8 = 11.52 xx 10^(-5) Wb//s`
As both the effects cause a decrese in magnetic flux along the positive Z-direction, therefore, they add up.
`:.` Total induced e.m.f. `e = 1.44 xx 10^(-5) + 11.52 xx 10^(-5) = 12.96 xx 10^(-5) V`
Induced current, `= (e)/(R ) = (13.96 xx 10^(-5))/(4.5 xx 10^(-3)) = 2.88 xx 10^(-2)`
The direction of induced current is such as to increase the flux through the loop along positive Z-direction. For example, for the observer, if the loop moves of the right, the current will be seen to be anticlockwise.
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