# A conducting wire XY of mass m and negligible resistance

Question:

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.

Solution:

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.