A thermocol vessel contains `0.5kg` of distilled water at `30^(@)C`. A metal coil of area `5 xx 10^(-3) m^(2)`, number of turns `100`, mass `0.06 kg` and resistance `1.6 Omega` is lying horizontally at the bottom of the vessel. A uniform time-varying magnetic field is set up to pass vertically through the coil at time `t = 0`. The field is first increased from zero to `0.8 T` at a constant rate between `0` and `0.2 s` and then decreased to zero at the same rate between `0.2 and 0.4s`. the cycle is repeated `12000` times. Make sketches of the current through the coil and the power dissipated in the coil as function of time for the first two cycles. Clearly indicate the magnitude of the quantities on the axes. Assumes that no heat is lost to the vessel or the surroundings. Determine the final tempreture of water under thermal equilibrium. Specific heat of metal ` = 500 j kg^(-1) K^(-1)` and the specific heat of water `= 4200 j kg^(-1) K^(-1)`. Neglect the inductance of coil.

Answer

`35.6^(@)C`

Explanation

The induced emf in the coil
`e=-N(dphi)/(dt)=-NA((dB)/(dt))`
Induced current `i=e/R=-(Na)/R((dB)/(dt))`.
Given `(dB)/(dt)=(0.8-0)/(0.2-0)=4T//s`
Thus for the time interval `i=(100(5xx10^(-3)))/1.6xx4` in the interval `0.2` to `0.4s, dB//dt` is negative and so induced current will be `+1.25 A`.
Power dissipated `P=i^(2)R=1.25^(2)xx1.6` The total energy dissipated in 12000 cycles `E=Pt=2.5xx(12000xx0.4)=12xx10^(3)J`

(a) Variation of current with time
(b) Variation of power with time
if `Deltat` is the change intemperature of water, then by convation of energy.
`12xx10^(3)=(m_(1)c_(1)+m_(2)c_(2))Deltat`
`:. Deltat=(12xx10^(3))/((m_(1)c_(1)+m_(2)c_(2)))`
`=(12xx10^(3))/((0.5xx4200+0.06xx500))=5.6^(@)C`
Final temperature `=T_(i)+Deltat=30+5.6=35.6^(@)C`