A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires. The closed-circuit has a resistance R due to AC.
AB and CD are perfect conductors. There is a magnetic field B = B t(k ).
(i) Write down the equation for the acceleration of the wire XY.
(ii) If B is independent of time, obtain v(t) , assuming v (0) = u0.
(iii) For (b), show that the decrease in kinetic energy of XY equals the heat lost in R.
The magnetic flux linked with the loop is given as ϕm = B.A = BA cos θ
Where A is the area vector and B is the magnetic field vector.
Emf induced due to change in magnetic field = e1 = -dB(t)/dt lx (t)
Emf induced due to motion = e2 = B(t) lv (t) (-j)
Total emf in the circuit = emf due to change in the field + the motional emf across XY
E = -dB(t)/dt lx (t) – B(t)lv(t)
We know current, I = E/R
Force = IB(t)/R[-dB(t)/dt Ix(t) – I2B2(t)/R dx/dt]
B is independent of the time
Power consumption is given as P = I2R
Therefore, energy consumed in time interval dt is given as = m/2 u02 – m/2 v2(t)
The above equation shows that there is a decrease in the kinetic energy.
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