A thermocol vessel contains 0.5 kg of distilled water at 30° C . A metal coil of area 5 × 10-3 m 2, number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform time varying magnetic field is setup to pass vertically through the coil at time t=0. The field is first increased from 0 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 0.4 s. The cycle is repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as a function of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axes. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium. Specific heat of metal =500 J / kg - K and the specific heat of water =4200 J / kg - K. Neglect the inductance of coil.

Question

A thermocol vessel contains 0.5 kg of distilled water at 30° C . A metal coil of area 5 × 10-3 m 2, number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform time varying magnetic field is setup to pass vertically through the coil at time t=0. The field is first increased from 0 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 0.4 s. The cycle is repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as a function of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axes. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium. Specific heat of metal =500 J / kg - K and the specific heat of water =4200 J / kg - K. Neglect the inductance of coil. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Magnetic field (B) varies with time (t) as shown in figure. | (dB/dt)| = (0.8/0.2) = 4 T/s Induced emf in the coil due to change in magnetic flux passing through it, e =| (d φ/dt)| = NA|(dB/dt)| Here, A = Area of coil = 5 × 10-3 m2 N = Number of turns =100 Substituting the values, we get e = (100) (5 × 10-3) (4)= 2 V Therefore, current passing through the coil i = (e/R) or i = (2/1.6) = 1.25 A NOTE That from 0 to 0.2 s and from 0.4 s (0.6 s, magnetic field passing through the coil increases, while during the time 0.2 s to 0.4 s and from 0.6 s to 0.8 s magnetic field passing through the coil decreases. Therefore, direction of current through the coil in these two time intervals will be opposite to each other. The variation of current (i) with time (t) will be follows: Power dissipated in the coil is P = i2R = (1.25)2 (1.6)= 2.5 W Power is independent of the direction of current through the coil. Therefore, power (P) versus time (t) graph for first two cycles will be as under: Total heat obtained in 12,000 cycles will be H = P.t = (2.5)( 12000)( 0.4) =12000 J This heat is used in raising the temperature of the coil and the water. Let 0 be the final temperature. Then H = mwsw(θ - 30) + mcsc(θ - 30) Here mw = mass of water = 0.5 kg sw = specific heat of water = 4200 J/kg-K mc = mass of coil = 0.06 kg and sc = specific heat of coil = 500 J/kg-K Substituting the values, we get 12000 = (0.5) (4200) (θ - 30) + (0.06) (500) (θ - 30) or θ = 35.6° C