Electromagnetic Induction - 7

M5. A coil of L = 50 x 10⁶ H and 5 is connected to a t = 5 V. A resistanc of 10 is connected parallel to the coil. Now at some instant the connection of the battery is switched off. Find the amount of heat generated in the coil after switching off the battery ?

Ans: Total energy stored in the inductor = LI₀²
E_L =
Fraction of energy lost across inductor.
          
          
          = 1.14 x 10^-4 J


H1. Two parrallel rails with negligable resistence 10 cm apart. They are connected by a 5 resistor. The circuit also contains two metal rods having resistances of 10 and 15 along the rails. The rods are pulled away from the resistor at constant speeds4 m/s and 2 m/s respectively.

A uniform magnetic field of magnitude .01 T is applied perpendicular to the plane of the rails. Determine the current in 5 resistor ?

Ans: In figure R = 5 , r₁ = 10 , r = 15 , e₁ = 4 x 10^-3 V, e₂ = 2 x 10^-3 V

The rod ab will act as a source of e.m.f.     e₁ = BV₁l₁ = .01 x 4 x .1
                                                                           = 4 x 10^-3
and     r₁ = 10

Similarly,     ef:     e₂ = BV₂l₂

                              = .01 x 2 x .1
                              = 2 x 10^-3 V
                           r₂ = 15


Refer figure (b)
Net resistance of the circuit = r₂ +
                                       = 15 +
                                       = A
Current I =
                = x 10^-3 A
Current through R,
   
Refer figure (c):
Net resistance = r₁ +
                     =
Current I' =
                 = x 10^-3 A
Current through R = I'
                             = mA
From superposition principal net current through 5 resistor is: I' - I = mA from d to c.


H2. A solenoid has an induction of 10 H and a resistance of 2 . It is connected to a 10 V battery. How long will ittake for the magnetic energy to reach 1/4 th of itsmaximum value.

Ans: Let the current through the circuit be I after a time t.
Then      + RI = V
or,        
integrating this expression keeping I = 0 at t = 0.
We get:      I = [1 - ]
               I = I₀(1 - ) ................................................. (1)
                 where I₀ = = Maximum Current.
Magnetic Enegy Stored = LI²
Here     LI² = (LI₀²)
or,        I =
Substituting the values in equation (1) we get:
         
or,      .5 = 1 -
or,      = .5
Which gives t = 3.478 seconds.


Key Words:

1. Magnetic Flux.
2. Faraday's Law.
3. Lenz's Law.
4. Motional E.M.F.
5. Self Induction.
6. Mutual Induction.
7. R - L Circuit.
8. L - C Oscillations.