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Home » 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.
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.
CBSE | Class 12 | Electromagnetic Induction | ncert exemplar | Physics
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.

m = B.A = BA cos m

The area vector is A, while the magnetic field vector is B.

e1 = -dB(t)/dt lx e1

e2 = B(t) lv (t)

The total emf in the circuit is equal to the emf owing to field change plus the motional emf across XY.

E = -dB(t)/dt lx (t) – B(t)lv

We are aware of current events. E/R = I

IB(t)/R[-dB(t)/dt Ix(t) – I2B2(t)/R dx/dt] Force = IB(t)/R[-dB(t)/dt Ix(t) – I2B2(t)/R dx/dt]

B is unaffected by the passage of time.

P = I2R is the formula for calculating power consumption.

As a result, the energy used during the time interval dt is calculated as = m/2 u02 – m/2 v2 (t)

The kinetic energy is decreasing, as shown by the equation above.

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